Subtopic Archive - Page 6 of 10

The Chain Rule of derivatives

September 27, 2019/by admin_ct

Derivative Using Chain Rule

This rule is used to find derivative of composite functions. In this section, we will study about differentiation of composition of two or more functions.

Lets assume that f(x) and g(x) are two differentiable functions, then their composition y= fog(x) will also be differentiable and its derivative is given as..

**Y’= f ’(g(x))*g’(x)**

Which can be read as derivative of outer function * derivative of inner function.

OR if we assume y=f(u) and u= g(x) then chain rule would be given as…

$\dpi{150} \mathbf{\mathbf{\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} u}*\frac{\mathrm{d}u }{\mathrm{d} x}}}$

Lets work on some examples to understand this rule better.

Example1: Find derivative of $\dpi{120} {\color{Red} y=(x^{2}+x+1)^{4}}$

Solution: Let $\dpi{120} y=u^{4}$ and $\dpi{120} u=(x^{2}+x+1)$

$\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} u} =4u^{3}$ $\dpi{120} \frac{\mathrm{d} u}{\mathrm{d} x}= 2x+1$

Applying chain rule $\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} u}*\frac{\mathrm{d} u}{\mathrm{d} x}$

$\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} x}= 4u^{3}*(2x+1)$

$\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} x}= 4(x^{2}+x+1)^{3}*(2x+1)$ [ answer]

Example2: Find derivative of y=log(secx+tanx)

Solution: let y= log(u) and u= secx+tanx

$\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} u}=\frac{1}{u}$ $\dpi{120} \frac{\mathrm{d} u}{\mathrm{d} x}= secx tanx +sec^{2}(x)$

Using chain rule,

$\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} u}*\frac{\mathrm{d} u}{\mathrm{d} x}$

$\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{u}*(secxtanx+sec^{2}x)$

$\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{secx+tanx}*[secx(tanx+secx)]$

$\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{secx(secx+tanx)}{secx+tanx}$

$\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} x}=secx$ [answer]

Example3: Find derivative of $\dpi{120} {\color{Red}Y=e^{tanx}}$

Solution: Using chain rule y’ = f ’(g(x))*g’(x)

Y= e^(tanx)

$\dpi{120} Y'=\left ( e^{tanx} \right )'*\left ( tanx \right )'$

$\dpi{120} Y'=\left ( e^{tanx} \right )*\left ( sec^{2}x\right )$

$\dpi{120} Y'= (sec^{2}x) e^{tanx}$ [answer]

Derivative of composition of three functions

Lets assume that f(x) ,g(x) and h(x) are three differentiable functions then their composition f(g(h(x)) is also differentiable.

Its derivative is given as…

**Y’ = f ’(g(h(x)))g’(h(x))h’(x)**

If we assume y= f(u) ,u= g(v) and v=h(x) then chain rule would be given as…

$\dpi{150} \mathbf{\frac{\mathrm{d} Y}{\mathrm{d} x}=\frac{\mathrm{d} Y}{\mathrm{d} u}*\frac{\mathrm{d} u}{\mathrm{d} v}*\frac{\mathrm{d} v}{\mathrm{d} x}}$

In this way we keep on taking derivative of layers until we reach innermost layer of composition function.

Example4: Find derivative of $\dpi{120} {\color{Red} Y = \sqrt{sin(3x^{2}+2x)}}$

Solution: Rewriting sqrt sign using exponents we get,

$\dpi{120} Y= \left ( sin(3x^{2}+2x) \right )^{\frac{1}{2}}$

$\dpi{120} Y'=\frac{1}{2}\left ( sin(3x^{2}+2x) \right )^{(\frac{1}{2}-1)}*\left ( sin(3x^{2}+2x) \right )'$

$\dpi{120} Y'=\frac{1}{2}\left ( sin(3x^{2}+2x) \right )^{\frac{-1}{2}}*\left ( cos(3x^{2}+2x) \right )*(3x^{2}+2x)'$

$\dpi{120} Y'=\frac{1}{2}\left ( sin(3x^{2}+2x) \right )^{\frac{-1}{2}}*\left ( cos(3x^{2}+2x) \right )*(6x+2)$

$\dpi{120} Y'=\frac{6x+2}{2}*\frac{1}{\sqrt{sin(3x^{2}+2x)}}*\left ( cos(3x^{2}+2x) \right )$

$\dpi{120} Y'=(3x+1)*\frac{cos(3x^{2}+2x)}{\sqrt{sin(3x^{2}+2x)}}$

$\dpi{120} Y'=\frac{(3x+1)cos(3x^{2}+2x)}{\sqrt{sin(3x^{2}+2x)}}$ [Answer]

Practice problems:

Find derivative of the following using chain rule.

1) y= log(cosx)

2) $\dpi{120} y=\sqrt{x^{2}+x+1}$

3) $\dpi{120} y=tan(e^{x}+5x)$

4) y= log(sin(3x+5))

5) $\dpi{120} y=e^{\sqrt{sinx}}$

Answers:

–tanx
$\dpi{120} \frac{2x+1}{2\sqrt{x^{2}+x+1}}$
$\dpi{120} (e^{x}+5)sec^{2}(e^{x}+5x)$
3cot(3x+5)
$\dpi{120} \frac{e^{\sqrt{sinx}}cosx}{2\sqrt{sinx}}$

The Product Rule for derivatives

September 27, 2019/by admin_ct

Derivative Using Product Rule

Product rule is used to find derivative of product of two functions. If f(x) and g(x) are two differentiable functions then product rule is defined as…

$\dpi{120} \mathbf{\frac{\mathrm{d} [f(x)*g(x)]}{\mathrm{d} x}=f(x)\frac{\mathrm{d} g(x)}{\mathrm{d} x}+g(x)\frac{\mathrm{d} f(x)}{\mathrm{d} x}}$

Using prime symbols this rule can be represented more beautifully.

[f(x)*g(x)]’= f(x)*g’(x) + g(x)*f ’(x)

In words it can be represented as…

Derivative of product of 2 functions=(1st function)*(derivative of 2^nd function) + ( 2^nd function)*(derivative of 1^st function)

Example 1: Find derivative of y=x sin(x)

Solution: Here lets assume f(x) = x and g(x) = sin(x)

f ’(X) =1 g’(x) = cos(x)

formula: [f(x)*g(x)]’= f(x)*g’(x) + g(x)*f’(x)

[x sin(x)]’ = x* cos(x) + sin(x)*1

= x cos(x)+ sin(x) [answer]

Example2: Find derivative of $\dpi{120} {\color{Red} y=x^{3}e^{x}}$

Solution: let f(x)= x^3 g(x)=e^x

f ’(x) = 3x^2 g’(x)= e^x

using values into the formula…

[f(x)*g(x)]’= f(x)*g’(x) + g(x)*f’(x)

$\dpi{120} \left ( x^{3}e^{x}\right )'= x^{3}e^{x}+e^{x}3x^{2}$

$\dpi{120} \left ( x^{3}e^{x}\right )'= x^{3}e^{x}+3x^{2}e^{x}$ [ answer]

Example3: Find derivative of y=(1+2tanx)(5+4cosx).

Solution: y = (1+2tanx)(5+4cosx)

Let f(x) = (1+2tanx) g(x) = (5+4cosx)

f ’(x) = 2 sec^2(x) g’(x)= -4sinx

Using product rule we get…

[(1+2tanx)(5+4cosx)]’ =(1+2tanx)(-4sinx)+( 2 sec^2(x))(5+4cosx)

= -4sinx(1+2tanx)+ 2sec^2(x) (5+4cosx)

= -4sinx -8sinxtanx +10sec^2(x)+8sec^2(x)cosx

= -4sinx -8sinxtanx +10sec^2x +8secx [answer]

This could also be solved other way, first multiplying the terms and then using fundamental rules to find their derivatives.

Y = 1(5+4cosx)+2tanx(5+4cosx)

Y = 5 + 4cosx + 10tanx +8tanxcosx

Y’ = 0 -4sinx +10sec^2(x)+ 8[tanx(cosx)’+cosx(tanx)’]

Y’ = -4sinx +10sec^2(x)+ 8tanx(-sinx)+8cosx sec^2(x)

Y’ = -4sinx +10sec^2(x) – 8tanx sinx + 8secx

Y’ = -4sinx -8sinxtanx +10sec^2(x) +8secx

We observe that answer is same using both ways!

Product rule for product of three terms

Let f(x), g(x) and h(x) be three differentiable functions. Then,

[f(x)*g(x)*h(x)]’= f ’(x)g(x)h(x) + f(x)g’(x)h(x) + f(x)g(x)h’(x)

We take derivative of each term one by one keeping other two same.

Same way we can find derivative of product of n terms taking derivative of each term one by one while keeping the others same.

Example4: Find derivative of $\dpi{120} {\color{Red} y=e^{x}x^{3}sinx}$

Solution: Let f(x)=e^x g(x)=x^3 h(x)= sinx

f’(x)=e^x g’(x)=3x^2 h’(x)= cosx

Using product rule for three terms….

Y’ = f ’(x)g(x)h(x) + f(x)g’(x)h(x) + f(x)g(x)h’(x)

$\dpi{120} y' = e^{x}x^{3}sinx +e^{x}3x^{2}sinx + e^{x}x^{3}cosx$

$\dpi{120} y' = e^{x}x^{3}sinx +3x^{2}e^{x}sinx + e^{x}x^{3}cosx$

$\dpi{120} y' = e^{x}x^{2}\left (xsinx +3sinx + xcosx \right )$

Practice problems:

Find derivative of the following using product rule:

Y = (x^2) cosx
Y = (3x+2)log(2x)
Y = sinx e^(3x)
Y =(x^4) log(x) tan(x)
$\dpi{120} y = \sqrt{x}e^{x}$

Answers:

x(2cosx-xsinx)
3log(2x) + 3+2/x
e^(3x) [ cosx +3sinx]
$\dpi{120} x^{3}[4logxtanx + tanx + xlogx sec^2(x)]$
$\dpi{120} \frac{e^{x}(2x+1)}{2\sqrt{x}}$

Derivative using limit definition

September 25, 2019/by admin_ct

Difference quotient :

Before moving to derivative using limit definition, it is good to be familiar with difference quotient.

Which is given as

$\dpi{150} \mathbf{\frac{f(x+h)-f(x)}{h}}$

This is used to compute slope of secant line passing through two points.

Example1: Find difference quotient for f(x)=5x-3

Solution:

$\dpi{120} \frac{f(x+h)-f(x)}{h}=\frac{\left (5(x+h)-3 \right )-(5x-3)}{h}$

$\dpi{120} \frac{\left (5x+5h-3 \right )-(5x-3)}{h}$

$\dpi{120} \frac{5x+5h-3-5x+3}{h}= \frac{5h}{h}=5$

Formula for derivative using limit definition.

$\dpi{150} \mathbf{f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}}$

Example2: Find derivative of f(x)= 3x^2+2x using limit definition.

Solution:

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{[3(x+h)^{2}+2(x+h)]-[3x^{2}+2x]}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{[3(x^{2}+2xh+h^{2})+2(x+h)]-[3x^{2}+2x]}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{[3x^{2}+6xh+3h^{2}+2x+2h]-[3x^{2}+2x]}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{3x^{2}+6xh+3h^{2}+2x+2h-3x^{2}-2x}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{6xh+3h^{2}+2h}{h}$ [ canceling terms on the numerator]

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{h(6x+3h+2)}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0} 6x+3h+2$

f'(x) = 6x+3(0)+2 = 6x+2

Example 3: Find derivative of f(x)= 5/x using limit definition.

Solution: $\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{\frac{5}{x+h}-\frac{5}{x}}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{\frac{5}{x+h}*\frac{x}{x}-\frac{5}{x}*\frac{x+h}{x+h}}{h}$ [ made the denominators same]

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{\frac{5x-5(x+h)}{x(x+h)}}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{\frac{5x-5x-5h}{x(x+h)}}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{\frac{-5h}{x(x+h)}}{h}=\lim_{h\rightarrow 0}\frac{-5h}{x(x+h)}*\frac{1}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{-5}{x(x+h)}$

$\dpi{120} f'(x)=\frac{-5}{x(x+0)}=\frac{-5}{x^{2}}$

Example4: Find derivative of f(x)=√(3x+5) using limit definition.

Solution:

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{\sqrt{3(x+h)+5}-\sqrt{3x+5}}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{\sqrt{3(x+h)+5}-\sqrt{3x+5}}{h}*\frac{\sqrt{3(x+h)+5}+\sqrt{3x+5}}{\sqrt{3(x+h)+5}+\sqrt{3x+5}}$ [rationalizing the numerator]

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{\left (\sqrt{3(x+h)+5} \right )^{2}-\left (\sqrt{3x+5} \right )^{2}}{h\left (\sqrt{3(x+h)+5}+\sqrt{3x+5} \right )}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{\left (3(x+h)+5 \right )-(3x+5) }{h\left (\sqrt{3(x+h)+5}+\sqrt{3x+5} \right )}$ [sqrt sign get eliminated with exponent 2]

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{3x+3h+5 -3x-5 }{h\left (\sqrt{3(x+h)+5}+\sqrt{3x+5} \right )}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{3h}{h\left (\sqrt{3(x+h)+5}+\sqrt{3x+5} \right )}$ [ simplified the numerator]

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{3}{\left (\sqrt{3(x+h)+5}+\sqrt{3x+5} \right )}$ [ canceled h from numerator and denominator]

$\dpi{120} f'(x)=\frac{3}{\left (\sqrt{3(x+0)+5}+\sqrt{3x+5} \right )}$

$\dpi{120} f'(x)=\frac{3}{\left (\sqrt{3x+5}+\sqrt{3x+5} \right )}=\frac{3}{2\sqrt{3x+5}}$

Example : Find derivative of f(x)= sin(x) using limit definition.

Solution:

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{sin(x+h)-sin(x)}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{2sin(\frac{x+h-x}{2})cos(\frac{x+h+x}{2})}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{2sin(\frac{h}{2})cos(\frac{2x+h}{2})}{h}$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{2sin(\frac{h}{2})}{h}\lim_{h\rightarrow 0}cos\left ( \frac{2x+h}{2} \right )$

$\dpi{120} f'(x)=\lim_{h\rightarrow 0}\frac{sin(\frac{h}{2})}{\frac{h}{2}}\lim_{h\rightarrow 0}cos\left ( x+\frac{h}{2} \right )$

f ‘(x) = (1) cos (x+0) formula : $\dpi{120} \mathbf{\lim_{x\rightarrow 0}\frac{sinx}{x}=1}$

f ‘(x) = cos(x)

Practice problems:

Find the derivative of following functions using limit definition.

f(x) = x^2 -5x
f(x) = 7/x
f(x) = √(x+2)
f(x) = cos x

Answers:

2x-5
$\dpi{120} \frac{-7}{x^{2}}$
$\dpi{120} \frac{1}{2\sqrt{x+2}}$
–sinx

Derivative using Fundamental rules

September 25, 2019/by admin_ct

Rules of derivative

lets be aware about some fundamental rules of differentiation.

Power rule $\dpi{120} \mathbf{\frac{\mathrm{d} (x)^{n}}{\mathrm{d} x}=nx^{n-1}}$

$\dpi{120} \mathbf{\frac{\mathrm{d} (x)}{\mathrm{d} x}=1}$ [derivative of x with respect to x is 1 always]

$\dpi{120} \mathbf{\frac{\mathrm{d} e^{kx}}{\mathrm{d} x}=ke^{kx}}$

$\dpi{120} \mathbf{\frac{\mathrm{d} (ln(kx))}{\mathrm{d} x}= \frac{1}{x}}$ [where k is any constant]

$\dpi{120} \mathbf{\frac{\mathrm{d} }{\mathrm{d} x}\left ( ln\left (\frac{k}{x} \right ) \right )=\frac{-1}{x}}$

$\dpi{120} \mathbf{\frac{\mathrm{d} \left (k f(x) \right )}{\mathrm{d} x}= k\frac{\mathrm{d} f(x)}{\mathrm{d} x}}$ [constant taken out while taking derivative]

$\dpi{120} \mathbf{\frac{\mathrm{d} \left ( a^{x} \right )}{\mathrm{d} x}= a^{x}\log_{e}a}$

To understand these rules better, lets work on some examples.

Example1 : Find derivative of y=x^3+2x+5

Solution: y=x^3+2x+5

To find derivative of an expression, we can take derivative of each term separately.

$\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} (x^{3})}{\mathrm{d} x}+2\frac{\mathrm{d} (x)}{\mathrm{d} x}+\frac{\mathrm{d} (5)}{\mathrm{d} x}$

$\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} x}=3x^{3-1}+2(1)+0$

$\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} x}=3x^{2}+2$ [answer]

Note : Derivative can also be represented using prime sign like y’ or f’(x) in addition to dy/dx

Example2 : Find derivative of

$\dpi{120} {\color{Red} y=x^{5}+e^{5x}}$

Solution: $\dpi{120} y'=\left (x^{5} \right )'+\left (e^{5x} \right )'$

$\dpi{120} y' = 5x^{5-1}+ 5e^{5x}$

$\dpi{120} y'= 5x^{4}+5e^{5x}$ [answer]

Example3: Find derivative of y= log(7x)

Solution: $\dpi{120} y'= \left ( log(7x) \right )'$

$\dpi{120} y'=\frac{1}{x}$ [ answer]

Derivative of log(kx) is always 1/x , no matter what the constant k is.

Example4: Find derivative of $\dpi{120} {\color{Red} y=3\sqrt{x}+\frac{5}{x}}$

Solution: $\dpi{120} y=3 x ^{\frac{1}{2}}+5x^{-1}$ [Rewrite the terms using exponents]

$\dpi{120} y'= 3*\frac{1}{2}x^{\frac{1}{2}-1}+5(-1x^{-1-1})$ [ Apply power rule ]

$\dpi{120} y'= \frac{3}{2}x^{\left (\frac{-1}{2} \right )}-5x^{-2}$

$\dpi{120} y'= \frac{3}{2\sqrt{x}}-\frac{5}{x^{2}}$ [answer]

Example5: Find derivative of

$\dpi{120} {\color{Red} y=5^{x}+ log(5x)+log\left (\frac{5}{x} \right )}$

Solution: $\dpi{120} y'=\left (5^{x} \right )'+ \left (log(5x) \right )'+\left (log\left (\frac{5}{x} \right ) \right )'$

$\dpi{120} y'=5^{x}\log_{e}5+ \frac{1}{x}+\frac{-1}{x}$ [using rules 4 ,5 and 7 given above]

$\dpi{120} y'= 5^{x}\log_{e}5$ [ answer]

Example6: Find derivative of

$\dpi{120} {\color{Red} y=\frac{3x^{2}+5x+2}{\sqrt{x}}}$

Solution: First, we should try to simplify the expression as much as possible.

$\dpi{120} y =\frac{3x^{2}}{\sqrt{x}}+\frac{5x}{\sqrt{x}}+\frac{2}{\sqrt{x}}$

$\dpi{120} y =\frac{3x^{2}}{x^{\frac{1}{2}}}+\frac{5x}{x^{\frac{1}{2}}}+\frac{2}{x^{\frac{1}{2}}}$

$\dpi{120} y=3x^{\left ( 2-\frac{1}{2} \right )}+5x^{\left (1-\frac{1}{2} \right )}+2x^{\frac{-1}{2}}$ [ simplified using rules of exponents]

$\dpi{120} y= 3x^{\frac{3}{2}}+5x^{\frac{1}{2}}+2x^{\frac{-1}{2}}$

$\dpi{120} y'= 3*\frac{3}{2}x^{\left (\frac{3}{2}-1 \right )}+5*\frac{1}{2}x^{\left (\frac{1}{2}-1 \right )}-2*(\frac{-1}{2})x^{\left (\frac{-1}{2} -1 \right )}$ [ applied formula 1(power rule) shown above]

$\dpi{120} y'= \frac{9}{2}x^{\frac{1}{2}}+\frac{5}{2}x^{\frac{-1}{2}}-x^{\frac{-3}{2}}$

$\dpi{120} y' =\frac{9}{2}\sqrt{x}+\frac{5}{2\sqrt{x}}-\frac{1}{x\sqrt{x}}$ [Answer]

Practice problems:

Find derivative of following functions using fundamental rules.

$\dpi{120} f(x)= \frac{3x^{2}+5x+4}{x}$

$\dpi{120} f(x)=e^{x}+\frac{5}{x}+ln|x|$

$\dpi{120} f(x)=\sqrt{x}(2x+3)$

f(x) =3^x+ log(3x) + log(3/x)

Answers:

$\dpi{120} 3-\frac{4}{x^{2}}$
$\dpi{120} e^{x}-\frac{5}{x^{2}}+\frac{1}{x}$
$\dpi{120} 3\sqrt{x}+\frac{3}{2\sqrt{x}}$
$\dpi{120} 3^{x}\log_{e}3+\frac{1}{x}-\frac{1}{x}$

Continuity

September 22, 2019/by admin_ct

What is Continuity

The meaning of term continuous is same as we use in our daily life. When we say that a function f(x) is continuous at point x=a, it means that at point (a,f(a)) graph of function has no holes or gaps. The graph is unbroken at point (a,f(a)).

A function is said to be continuous at a point x=a iff

f(a) is defined
$\dpi{120} \mathbf{\lim_{x\rightarrow a}f(x)}$ exist
$\dpi{120} \mathbf{\lim_{x\rightarrow a}f(x)= f(a)}$

Any function f(x) fails to be continuous at x=a if it fail in any of three conditions given above and then it is called discontinuous function.

A function is continuous on interval [a,b] if it is continuous at each point of this interval.

Lets understand this concept of continuity and discontinuity with the help of graphs.

A function f(x) is said to be left continuous (continuous from left) at x=a, iff

$\dpi{120} \lim_{x\rightarrow a^{-}}f(x)$ exist and $\dpi{120} \lim_{x\rightarrow a^{-}}f(x)= f(a)$

A function f(x) is said to be right continuous(continuous from right) at x=a, iff

$\dpi{120} \lim_{x\rightarrow a^{+}}f(x)$ exist and $\dpi{120} \lim_{x\rightarrow a^{+}}f(x)= f(a)$

It follows from the above definition that f(x) is continuous at x=a iff it is both left and right continuous at x=a.

$\dpi{150} \mathbf{\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)= f(a)}$

Algebra of continuous functions:

Let f and g be two real functions, continuous at x=a, then sum (f+g), difference (f-g) , product (f*g) and quotient (f/g) of these continuous functions will also be continuous at x=a.

In the next example lets learn how a graph can be discontinuous at different points and what it mean for a function to be continuous.

Example1 : Given the graph of f(x) shown below, check if f(x) is continuous at x=-2, x=0 and x=3.

Solution: At x= -2

We see that graph is left continuous at x=-2, That is

$\dpi{120} \lim_{x\rightarrow 2^{-}}f(x)=2$ and f(-2)=2

But it is not right continuous , because

$\dpi{120} \lim_{x\rightarrow 2^{+}}f(x)= -1$ and f(-2)=2

Since limit doesn’t exist at x=-2 so this is discontinuous at x=-2 and this type of discontinuity is called jump discontinuity.

At x= 0,

Clearly graph is connected and smooth at x=0 . We can also see that both left and right sided limits are same as function value at x=0

$\dpi{120} \lim_{x\rightarrow 0}f(x)= f(0)= 1$
So graph is continuous at x=0

At x= 3,

We can see that limit is approaching to same value from either (left or right) sides. That is

$\dpi{120} \lim_{x\rightarrow 3^{-}}f(x)=\lim_{x\rightarrow 3^{+}}f(x)= 0$

But f(3) = -1

That means $\dpi{120} \lim_{x\rightarrow 3}f(x) \neq f(3)$

So functions is discontinuous at x=3. This is called removable discontinuity.

Example2 : Test the continuity of given function at origin

$\dpi{120} {\color{Red} f(x)= \left\{\begin{matrix} \frac{|x|}{x} & x\neq 0\\ 1 & x=0 \end{matrix}\right.}$

Solution: Using the definition of absolute function we have,

$\dpi{120} |x|= \left\{\begin{matrix} x & x \geq 0\\ -x & x< 0 \end{matrix}\right.$

Using this definition we get the new transformed function as,

$\dpi{120} f(x)=\left\{\begin{matrix} \frac{x}{x} & x> 0 \\ \frac{-x}{x} & x< 0 \\ 1& x=0 \end{matrix}\right.$

$\dpi{120} f(x)=\left\{\begin{matrix} 1 & x> 0 \\ -1 & x< 0 \\ 1& x=0 \end{matrix}\right.$

We have f(x) =1 for domain x >0 that means right sided limit at x=0 is 1

$\dpi{120} \lim_{x\rightarrow 0^{+}}f(x)=\lim_{x\rightarrow 0}1=1$

We have f(x) =-1 for domain x <0 that means left sided limit at x=0 is -1

$\dpi{120} \lim_{x\rightarrow 0^{-}}f(x)=\lim_{x\rightarrow 0}(-1)=-1$

Since left and right sided limits are not same,

$\dpi{120} \lim_{x\rightarrow 0^{+}}f(x)\neq \lim_{x\rightarrow 0^{-}}f(x)$

So limit doesn’t exist at x=0 and therefore function is not continuous at x=0.

Example3: Consider the function

$\dpi{120} {\color{Red} f(x)=\left\{\begin{matrix} x^{2}-3x &x< -1 \\ 2x-3 & -1\leq x< 1\\ 2 & x=1\\ \frac{1}{x-2} & x\geq 1 \end{matrix}\right.}$

Evaluate the following expressions. When a limit fails to exist, cite the reason why this is the case. Then identify all discontinuities present.

f(-1)
f(1)
$\dpi{120} {\color{Red} \lim_{x\rightarrow -1}f(x)}$
$\dpi{120} {\color{Red} \lim_{x\rightarrow 0}f(x)}$
$\dpi{120} {\color{Red} \lim_{x\rightarrow 1}f(x)}$
$\dpi{120} {\color{Red} \lim_{x\rightarrow 2}f(x)}$

Solution 1):

Since x = -1 lie in the domain -1 ≤ x < 0 so we consider function f(x)= 2x-3 to find f(-1)

f(-1) = 2(-1)-3 = -5

Given f(1)= 2

We need to check left and right sided limit at x=-1. For that we choose respective domains.

left sided limit:

$\dpi{120} \lim_{x\rightarrow -1^{-}}f(x)=\lim_{x\rightarrow -1}x^{2}-3x = (-1)^{2}-3(-1)=4$

Right sided limit:

$\dpi{120} \lim_{x\rightarrow -1^{+}}f(x)=\lim_{x\rightarrow -1}2x-3 =2(-1)-3 =-5$

Since left and right sided limits are not same so limit doesn’t exist at x=-1.

x=0 lie in the domain -1 x < 1 and function given for this domain is f(x)=2x-3

$\dpi{120} \lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}2x-3 =2(0)-3 =-3$

For this too we find left and right sided limits at x=1

Left sided limit:

$\dpi{120} \lim_{x\rightarrow 1^{-}}f(x)=\lim_{x\rightarrow 1}2x-3 = 2(1)-3=-1$

Right sided limit:

$\dpi{120} \lim_{x\rightarrow 1^{+}}f(x)=\lim_{x\rightarrow 1}\frac{1}{x-2} = \frac{1}{1-2} =-1$

Left and right sided limits are same so limit exist

$\dpi{120} \lim_{x\rightarrow 1}f(x)=-1$

6) $\dpi{120} \lim_{x\rightarrow 2}f(x)= \lim_{x\rightarrow 2}\frac{1}{x-2}$

We see that x=2 is a vertical asymptote here. Lets check left and right sided limits at x=2

Lets assume a very small quantity ‘h’ which tends to 0 and can be used judiciously for left and right sided limits.

Left sided limit:

$\dpi{120} \lim_{x\rightarrow 2^{-}}f(x)=\lim_{h\rightarrow 0}f(2-h)= \lim_{h\rightarrow 0}\frac{1}{2-h-2} = \frac{-1}{h}= -\infty$

Right sided limit:

$\dpi{120} \lim_{x\rightarrow 2^{+}}f(x)=\lim_{h\rightarrow 0}f(2+h)= \lim_{h\rightarrow 0}\frac{1}{2+h-2} = \frac{1}{h}= \infty$

Since left and right limits are not same at x=2 so limit doesn’t exist at x=2

Lets discuss discontinuities now. Clearly function is not continuous where limit doesn’t exist. So this function is not continuous at x=-1 and x=2.

Limit exist at x=0 and linear functions are always continuous in their domains. So this function is continuous at x=0.

We see that limit exist at x=1 but it is not same as the function value at x=1

$\dpi{120} \lim_{x\rightarrow 1^{-}}f(x)=\lim_{x\rightarrow 1^{+}}f(x)=-1$ But f(1)= 2

Which shows there is a hole and so the function is not continuous here.

Example4 : If $\dpi{120} {\color{Red} f(x)=\left\{\begin{matrix} 1 & x\leq 3 \\ ax+b& 3< x< 5 \\ 7& 5\leq x \end{matrix}\right.}$

Determine the values of a and b so that f(x) is continuous.

Solution: Given function is a constant function for all x<3 and for all x>5 so it is continuous for all x values in these domains. For domain 3<x<5 given function is linear function and linear function is always continuous in its domain. Therefore f(x) is continuous at each x∈R except possibly at x=3 and x=5.

At x=3,

Left limit:

$\dpi{120} \lim_{x\rightarrow 3^{-}}f(x)= \lim_{x\rightarrow 3}(1) = 1$

and f(3) =1

Right Limit:

$\dpi{120} \lim_{x\rightarrow 3^{+}}f(x)= \lim_{x\rightarrow 3}(ax+b) = 3a+b$

For f(x) to be continuous at x=3, we must have

$\dpi{120} \lim_{x\rightarrow 3^{-}}f(x)=\lim_{x\rightarrow 3^{+}}f(x)= f(3)$

1 = 3a+b ——-(i)

At x=5,

Left limit:

$\dpi{120} \lim_{x\rightarrow 5^{-}}f(x)= \lim_{x\rightarrow 5}(ax+b) = 5a+b$

Right limit:

$\dpi{120} \lim_{x\rightarrow 5^{+}}f(x)= \lim_{x\rightarrow 5}(7) = 7$

and f(5) =7

For f(x) to be continuous at x=5, we must have

$\dpi{120} \lim_{x\rightarrow 5^{-}}f(x)=\lim_{x\rightarrow 5^{+}}f(x)= f(5)$

5a+b=7 ——-(ii)

Solving equations (i) and (ii)

3a+b =1

5a+b=7

we get, a=3 and b=-8

Practice problems:

1) A function f(x) defined as $\dpi{120} f(x)= \left\{\begin{matrix} \frac{x^{2}-9}{x-3} & x\neq 3\\ 6 & x=3 \end{matrix}\right.$

Show that f(x) is continuous at x=3.

2) Consider the function

$\dpi{120} f(x)=\left\{\begin{matrix} \frac{2x+3}{x+2} & x< 1 \\ x^{2} &1\leq x< 2 \\ 5x-6 & x\geq 2 \end{matrix}\right.$

Evaluate the following expressions. When a limit fails to exist, cite the reason why this is the case. Then identify all discontinuities present.

f(2)

$\dpi{120} \lim_{x\rightarrow -2}f(x)$

$\dpi{120} \lim_{x\rightarrow 1}f(x)$

$\dpi{120} \lim_{x\rightarrow 2}f(x)$

3 ) Determine the value of constant k so that the function

$\dpi{120} f(x)= \left\{\begin{matrix} \frac{x^{2}-9}{x-3} & x\neq 3\\ 6 & x=3 \end{matrix}\right.$

is continuous at x=0.

Answers:

2) f(2)=4 , Limit doesn’t exist at x=-2,1 ; $\dpi{120} \lim_{x\rightarrow 2}f(x)=4$

3) 2/5

Indeterminate forms and L’Hospital’s rule

September 22, 2019/by admin_ct

Indeterminate forms

Sometimes we plug in limit and get the result as 0/0 or ∞/∞ . Which are called indeterminate forms. Other indeterminate forms are

$\dpi{150} \mathbf{1^{\infty },\infty ^{\infty },\infty ^{0},\infty -\infty ,0*\infty }$

Whenever we face such situations, we use L’Hospital’s rule. As per this rule we take derivative of numerator and denominator separately like $\dpi{120} \mathbf{\frac{f'(x)}{g'(x)}}$ .This rule work best for 0/0 and ∞/∞ forms. If we get any other indeterminate forms than these two, we change them into quotient so that we can apply L’Hospital’s rule.

Sometimes some expressions like $\dpi{120} \mathbf{\lim_{x\rightarrow 4}\frac{x^{2}-16}{x-4}}$ resulting in 0/0 or ∞/∞ forms can be simplified using factoring etc. to get a defined value but it is not always the case. Then L’Hospital’s rule comes for our rescue.

Lets work on some examples to know this rule better.

Example1 : Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow \infty }\frac{e^{x}}{x^{2}}}$

Solution: Here, if we plug in limit x as ∞ we get ∞/∞ , which is indeterminate form. So we have to use L’Hospital’s rule.

Using that rule we take derivative of numerator and denominator separately. Note that here we can’t use quotient rule to find derivative .

$\dpi{120} \lim_{x\rightarrow \infty }\frac{e^{x}}{x^{2}}=\lim_{x\rightarrow \infty }\frac{(e^{x})'}{(x^{2})'}= \lim_{x\rightarrow \infty }\frac{e^{x}}{2x}$

If we plug in limit x as ∞ , again we get ∞/∞ indeterminate form. That mean we have to use the same rule second time.

$\dpi{120} \lim_{x\rightarrow \infty }\frac{e^{x}}{2x}=\lim_{x\rightarrow \infty }\frac{(e^{x})'}{(2x)'}= \lim_{x\rightarrow \infty }\frac{e^{x}}{2}=\frac{\infty }{2}=\infty$ [answer]

When indeterminate form is different than 0/0 or ∞/∞ :

Then we have to rewrite the function as a quotient to be able to apply L’Hospital’s rule.

Example2 : Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow -\infty }xe^{x}}$

Solution: When we plug in limit, we get indeterminate form as -∞*0 . So we have to change the function as quotient so that we can apply L’Hospital’s rule.

$\dpi{120} \lim_{x\rightarrow -\infty }x e^{x} =\lim_{x\rightarrow -\infty }\frac{x}{e^{-x }}$

$\dpi{120} \lim_{x\rightarrow -\infty }\frac{(x)'}{\left (e^{- x} \right )'}= \lim_{x\rightarrow -\infty }\frac{1}{-e^{-x}}$

$\dpi{120} \frac{1}{-e^{\infty }}= \frac{1}{-\infty }=0$

Note: Be careful about choosing numerator and denominator. If we choose e^x as numerator and 1/x as denominator then this will end up nowhere .

Limit using ln :

Limit of exponential functions of the form $\dpi{120} f(x)^{g(x)}$ can be found easily taking ln of both sides. Here is an example.

Example3 : Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow \infty }(x)^{\frac{1}{x}}}$

Solution: Let y = $\dpi{120} \lim_{x\rightarrow \infty }(x)^{\frac{1}{x}}$

Taking ln of both sides ,

ln y = ln $\dpi{120} \left (\lim_{x\rightarrow \infty }(x)^{\frac{1}{x}} \right )$

lny = $\dpi{120} \lim_{x\rightarrow \infty } \left ( lnx^{\frac{1}{x}} \right )$

ln y = $\dpi{120} \lim_{x\rightarrow \infty }\frac{1}{x}lnx$

lny = $\dpi{120} \lim_{x\rightarrow \infty }\frac{\left (lnx \right )'}{(x)'}$ [ using L’Hospital’s rule because of

ln y = $\dpi{120} \lim_{x\rightarrow \infty }\frac{\left (\frac{1}{x}\right )}{1}$

lny = $\dpi{120} \lim_{x\rightarrow \infty }\frac{1}{x} = \frac{1}{\infty }=0$

lny = 0

$\dpi{120} y = e^{0} =1$ [answer]

Example4: Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow \infty }\left ( 1+\frac{1}{x} \right )^{x}}$

Solution: let y = $\dpi{120} \lim_{x\rightarrow \infty }\left ( 1+\frac{1}{x} \right )^{x}$

Taking ln of both sides,

lny = ln $\dpi{120} \lim_{x\rightarrow \infty }\left ( 1+\frac{1}{x} \right )^{x}$

lny = $\dpi{120} \lim_{x\rightarrow \infty }ln\left ( 1+\frac{1}{x} \right )^{x}$

lny = $\dpi{120} \lim_{x\rightarrow \infty }x ln\left ( 1+\frac{1}{x} \right )$

Rewriting the expression as a quotient,

ln y = $\dpi{120} \lim_{x\rightarrow \infty }\frac{ln\left ( 1+\frac{1}{x} \right )}{\frac{1}{x}}$

Using L’Hospital’s rule,

ln y = $\dpi{120} \lim_{x\rightarrow \infty }\frac{\left (ln\left ( 1+\frac{1}{x} \right ) \right )'}{\left (\frac{1}{x} \right )'}$

lny = $\dpi{120} \lim_{x\rightarrow \infty }\frac{\left (ln(x+1)-lnx \right )'}{\left (\frac {1}{x}\right )' }$

lny = $\dpi{120} \lim_{x\rightarrow \infty }\frac{\frac{1}{x+1}-\frac{1}{x}}{-x^{-2}}$

lny = $\dpi{120} \lim_{x\rightarrow \infty }\frac{\frac{x-x-1}{x(x+1)}}{\frac{-1}{x^{2}}}$

lny = $\dpi{120} \lim_{x\rightarrow \infty }\frac{\frac{-1}{x^{2}+x}}{\frac{-1}{x^{2}}}$

lny = $\dpi{120} \lim_{x\rightarrow \infty }\frac{-1}{x^{2}+x}*\frac{-x^{2}}{1}$

lny = $\dpi{120} \lim_{x\rightarrow \infty }\frac{x^{2}}{x^{2}+x}$

lny = $\dpi{120} \lim_{x\rightarrow \infty }\frac{x^{2}}{x^{2}(1+\frac{1}{x})}$

lny = $\dpi{120} \lim_{x\rightarrow \infty }\frac{1}{(1+\frac{1}{x})}$

lny = $\dpi{120} \frac{1}{1+\frac{1}{\infty }}= \frac{1}{1+0}=1$

lny = 1

y = e^1 = e [answer]

Practice problems:

Evaluate the following limits

$\dpi{120} \lim_{x\rightarrow \infty }\frac{e^{x^{2}}-cosx}{x^{2}}$

$\dpi{120} \lim_{x\rightarrow \infty }(sinx)^{x}$

$\dpi{120} \lim_{x\rightarrow \infty }\frac{4^{x}-5^{x}}{x}$

$\dpi{120} \lim_{x\rightarrow \infty }(cos6x)^{\frac{1}{x^{2}}}$

Answers:

3/2
1
ln(4/5)
e^(-18)

Exponential and logarithmic limits

September 20, 2019/by admin_ct

Limits of Exponential and logarithmic functions :

To evaluate the limits of exponential and logarithmic functions, following formulas are very useful.

$\dpi{150} \mathbf{\lim_{x\rightarrow 0}\frac{a^{x}-1}{x}= \log_{e}a = lna}$

$\dpi{150} \mathbf{\lim_{x\rightarrow 0}\frac{e^{x}-1}{x}= \log_{e}e = 1}$

$\dpi{150} \mathbf{\lim_{x\rightarrow 0}(1+kx)^{\frac{1}{x}}= e^{k}}$

$\dpi{150} \mathbf{\lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}= e}$

$\dpi{150} \mathbf{\lim_{x\rightarrow \infty }\left (1+\frac{1}{x} \right )^{x}= e}$

$\dpi{150} \mathbf{\lim_{x\rightarrow \infty }\left (1+\frac{k}{x} \right )^{x}= e^{k}}$

$\dpi{150} \mathbf{\lim_{x\rightarrow 0}\frac{ln(1+x)}{x}=1}$

Following examples will illustrate the applications of above results.

Example1 : Evaluate

$\dpi{150} {\color{Red} \lim_{x\rightarrow 0}\frac{a^{x}-b^{x}}{sinx}}$

Solution:

$\dpi{150} \lim_{x\rightarrow 0}\frac{a^{x}-b^{x}}{sinx}=\lim_{x\rightarrow 0}\frac{a^{x}-b^{x}-1+1}{sinx}$

$\dpi{150} \lim_{x\rightarrow 0}\frac{\left (a^{x}-1 \right )-\left (b^{x}-1 \right )}{sinx}$

$\dpi{150} \lim_{x\rightarrow 0}\frac{a^{x}-1}{sinx}-\frac{b^{x}-1}{sinx}$

$\dpi{150} \lim_{x\rightarrow 0}\frac{a^{x}-1}{sinx}-\lim_{x\rightarrow 0}\frac{b^{x}-1}{sinx}$

$\dpi{150} \lim_{x\rightarrow 0}\frac{a^{x}-1}{x}*\frac{x}{sinx}-\lim_{x\rightarrow 0}\frac{b^{x}-1}{x}*\frac{x}{sinx}$

$\dpi{150} \lim_{x\rightarrow 0}\frac{a^{x}-1}{x}*\lim_{x\rightarrow 0}\frac{x}{sinx}-\lim_{x\rightarrow 0}\frac{b^{x}-1}{x}*\lim_{x\rightarrow 0}\frac{x}{sinx}$

$\dpi{150} ln(a)*1 - ln(b)*1$

$\dpi{150} ln(a)-ln(b)=ln\left ( \frac{a}{b} \right )$

Example2: Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow 0}\frac{3^{x}+3^{-x}-2}{x^{2}}}$

Solution:

$\dpi{120} \lim_{x\rightarrow 0}\frac{3^{x}+3^{-x}-2}{x^{2}}=\lim_{x\rightarrow 0}\frac{3^{2x}+1-2(3^{x})}{x^{2}3^{x}}$

$\dpi{120} \lim_{x\rightarrow 0}\frac{3^{2x}-2(3^{x})+1}{x^{2}3^{x}}=\lim_{x\rightarrow 0}\frac{(3^{x}-1)^{2}}{x^{2}3^{x}}$

$\dpi{120} \lim_{x\rightarrow 0}\frac{(3^{x}-1)^{2}}{x^{2}}\lim_{x\rightarrow 0}\frac{1}{3^{x}}$

$\dpi{120} \lim_{x\rightarrow 0}\left (\frac{3^{x}-1}{x} \right )^{2}\lim_{x\rightarrow 0}\frac{1}{3^{x}}$

$\dpi{120} \left ( ln3 \right )^{2}\frac{1}{3^{0}}= (ln3)^{2}$

Example: Prove that

$\dpi{120} {\color{Red} \lim_{x\rightarrow 2}\frac{3^{x}+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}}=\frac{-4}{3}}$

Solution:

$\dpi{120} \lim_{x\rightarrow 2}\frac{3^{x}+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}}=\lim_{x\rightarrow 2}\frac{3^{x}+3^{3}3^{-x}-12}{3^{3}3^{-x}-3^{\frac{x}{2}}}$

$\dpi{120} \lim_{x\rightarrow 2}\frac{3^{2x}+3^{3}-12(3^{x})}{3^{3}-3^{\frac{x}{2}}3^{x}}$ [multiply top and bottom with 3^x ]

$\dpi{120} \lim_{x\rightarrow 2}\frac{3^{2x}+ 27-12(3^{x})}{27-3^{\frac{3x}{2}}}$

Lets use following substitution used for factoring the expression

$\dpi{120} 3^{x}=y ,3^{2x}=y^{2} and , 3^{\frac{x}{2}}=\sqrt{y}$

$\dpi{120} \lim_{x\rightarrow 2}\frac{y^{2}+ 27-12y}{27-y^{\frac{3}{2}}}=\lim_{x\rightarrow 2}\frac{y^{2}-12y+27}{27-(\sqrt{y})^{3}}$

$\dpi{120} \lim_{x\rightarrow 2}\frac{(y-9)(y-3)}{(3-\sqrt{y})(3^{2}+3\sqrt{y}+y)}$

$\dpi{120} \lim_{x\rightarrow 2}\frac{(\sqrt{y}-3)(\sqrt{y}+3)(y-3)}{(3-\sqrt{y})(3^{2}+3\sqrt{y}+y)}$

$\dpi{120} \lim_{x\rightarrow 2}\frac{-(3-\sqrt{y})(3+\sqrt{y})(y-3)}{(3-\sqrt{y})(9+3\sqrt{y}+y)}$

$\dpi{120} \lim_{x\rightarrow 2}\frac{-(3+\sqrt{y})(y-3)}{(9+3\sqrt{y}+y)}$

Plugin back the expression in terms of x, we get

$\dpi{120} \lim_{x\rightarrow 2}\frac{-(3+3^{\frac{x}{2}})(3^{x}-3)}{(9+3*3^{\frac{x}{2}}+3^{})}$

Apply limit as x=2

$\dpi{120} \frac{-(3+3^{\frac{2}{2}})(3^{2}-3)}{9+3*3^{\frac{2}{2}}+3^{2}}$

$\dpi{120} \frac{-(3+3)(9-3)}{9+3*3+9}=\frac{-36}{27}=\frac{-4}{3}$

Practice problems:

Evaluate the following limits:

$\dpi{120} \lim_{x\rightarrow 0}\frac{3^{x}-2^{x}}{tanx}$

$\dpi{120} \lim_{x\rightarrow 0}\frac{e^{x}+e^{-x}-2}{x^{2}}$

$\dpi{120} \lim_{x\rightarrow 0}\frac{x(e^{x}-1)}{1-cosx}$

Answers:

ln(3/2)
1
2

Limits of Trigonometric functions

September 20, 2019/by admin_ct

How to find limit of trigonometric functions.

When variable x tends to 0: Then we use the following basic limit formulas.

$\dpi{150} \mathbf{\lim_{x\rightarrow 0}\frac{sin(x)}{x}=1}$

$\dpi{150} \mathbf{\lim_{x\rightarrow 0}\frac{tan(x)}{x}=1}$

Lets work on some examples .

Example1 . Evaluate $\dpi{120} {\color{Red} \lim_{x\rightarrow 0}\frac{sin^{2}(3x)}{x^{2}}}$

Solution:

$\dpi{120} \lim_{x\rightarrow 0}\frac{sin^{2}(3x)}{x^{2}}=\lim_{x\rightarrow 0}\frac{sin(3x)sin(3x)}{x.x}$

$\dpi{120} \lim_{x\rightarrow 0}\frac{sin(3x)}{x}*\frac{sin(3x)}{x}$

$\dpi{120} \lim_{x\rightarrow 0}\frac{sin(3x)}{x}*\lim_{x\rightarrow 0}\frac{sin(3x)}{x}$ [using properties of limits]

$\dpi{120} \lim_{x\rightarrow 0}3\frac{sin(3x)}{3x}*\lim_{x\rightarrow 0}3\frac{sin(3x)}{3x}$

$\dpi{120} 3\lim_{x\rightarrow 0}\frac{sin(3x)}{3x}*3\lim_{x\rightarrow 0}\frac{sin(3x)}{3x}$
= 3(1) * 3(1) = 9

Example 2: Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow 0}\frac{1-cos\left ( mx \right )}{1-cos\left ( nx \right )}}$

Solution: Using half angle identity,

$\dpi{120} \lim_{x\rightarrow 0}\frac{1-cos\left ( mx \right )}{1-cos\left ( nx \right )}=\lim_{x\rightarrow 0}\frac{2sin^{2}\left ( \frac{mx}{2} \right )}{2sin^{2}\left ( \frac{nx}{2} \right )}$

$\dpi{120} \lim_{x\rightarrow 0}\left ( \frac{sin\left ( \frac{mx}{2} \right )}{sin\left ( \frac{nx}{2} \right )} \right )^{2}$

$\dpi{120} \lim_{x\rightarrow 0}\left ( \frac{\frac{sin(\frac{mx}{2})\frac{mx}{2}}{\frac{mx}{2}}}{\frac{sin\left ( \frac{nx}{2} \right )\frac{nx}{2}}{\frac{nx}{2}}} \right )^{2}$

$\dpi{120} \left (\frac{mx}{nx}\lim_{x\rightarrow 0}\frac{\frac{sin(\frac{mx}{2})}{\frac{mx}{2}}}{\frac{sin(\frac{nx}{2})}{\frac{nx}{2}}} \right )^{2}$

$\dpi{120} \frac{m^{2}}{n^{2}}\left (\frac{\lim_{x\rightarrow 0\frac{sin(\frac{mx}{2}}{\frac{mx}{2}})}}{\lim_{x\rightarrow 0\frac{sin(\frac{nx}{2}}{\frac{nx}{2}})}} \right )^{2}$

$\dpi{120} \frac{m^{2}}{n^{2}}\left ( \frac{1}{1} \right )^{2} = \frac{m^{2}}{n^{2}}$

Example3: Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow \infty }x tan\left ( \frac{1}{x} \right )}$

Solution: Here we use change of variables method.

Let $\dpi{120} \frac{1}{x}=y$ then as x–>∞ , y–>0

And x=1/y

We get the new expression as,

$\dpi{120} \lim_{x\rightarrow \infty }x tan\left ( \frac{1}{x} \right )= \lim_{y\rightarrow 0}\frac{1}{y}tan(y)$

$\dpi{120} \lim_{y\rightarrow 0}\frac{tany}{y} =1$

When variable x tends to non zero quantity:

In that case trigonometric limits can be evaluated by factorization . Lets work on some similar examples.

$\dpi{120} {\color{Red} \lim_{x\rightarrow \pi }\frac{1+sec^{3}x}{tan^{2}x}}$

Solution:

$\dpi{120} \lim_{x\rightarrow \pi }\frac{1+sec^{3}x}{tan^{2}x}= \lim_{x\rightarrow \pi }\frac{1+sec^{3}x}{sec^{2}x-1}$

$\dpi{120} \lim_{x\rightarrow \pi }\frac{(secx+1)(1+sec^{2}x-secx)}{(secx+1)(secx-1)}$

$\dpi{120} \lim_{x\rightarrow \pi }\frac{(1+sec^{2}x-secx)}{(secx-1)}$

$\dpi{120} \frac{1+(-1)^{2}-(-1)}{-1-1}= \frac{1+1+1}{-2}= \frac{-3}{2}$ [sec(π) =-1]

Example5 : Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow \frac{\pi }{2}}\frac{\sqrt{2}-\sqrt{1+sinx}}{\sqrt{2}cos^{2}x}}$

Solution:

$\dpi{120} \lim_{x\rightarrow \frac{\pi }{2}}\frac{\sqrt{2}-\sqrt{1+sinx}}{\sqrt{2}cos^{2}x}*\frac{\sqrt{2}+\sqrt{1+sinx}}{\sqrt{2}+\sqrt{1+sinx}}$

$\dpi{120} \lim_{x\rightarrow \frac{\pi }{2}}\frac{\left (\sqrt{2} \right )^{2}-\left (\sqrt{1+sinx} \right )^{2}}{\sqrt{2}cos^{2}x(\sqrt{2}+\sqrt{1+sinx})}$

$\dpi{120} \lim_{x\rightarrow \frac{\pi }{2}}\frac{{2} -\left (1+sinx \right )}{\sqrt{2}(1-sin^{2}x)(\sqrt{2}+\sqrt{1+sinx})}$

$\dpi{120} \lim_{x\rightarrow \frac{\pi }{2}}\frac{ 1-sinx}{\sqrt{2}(1-sinx)(1+sinx)(\sqrt{2}+\sqrt{1+sinx})}$

$\dpi{120} \lim_{x\rightarrow \frac{\pi }{2}}\frac{ 1}{\sqrt{2}(1+sinx)(\sqrt{2}+\sqrt{1+sinx})}$

$\dpi{120} \frac{1}{\sqrt{2}(1+1)(\sqrt{2}+\sqrt{2})}=\frac{1}{2\sqrt{2}(2\sqrt{2})}=\frac{1}{8}$

Practice problems:

Evaluate the following limits.

$\dpi{120} \lim_{x\rightarrow 0}\frac{1-cosx}{x^{2}}$
$\dpi{120} \lim_{x\rightarrow 0}\frac{sin(3x)+7x}{4x+sin(2x)}$ Hint: divide both top and bottom by x
$\dpi{120} \lim_{x\rightarrow 0}\frac{\sqrt{2}-\sqrt{1+cosx}}{sin^{2}x}$
$\dpi{120} \lim_{x\rightarrow \frac{3\pi }{2}}\frac{1+ csc^{3}x}{cot^{2}x}$

Answers:

1. 1/2

2. 5/3

3. 1/(4√2)

4. -3/2

Algebraic limits at infinity

September 19, 2019/by admin_ct

How to find algebraic limits at infinity.

Algebraic limits at infinity

Different functions behave differently when x values approach to infinity. For example polynomial functions behave differently than rational functions when x approach to infinity. Lets discuss them one by one.

Limit at infinity of Polynomial functions:

Limit of polynomials at infinity depend on their degree. Degree is the highest exponent of function or the exponent of leading term when written in order.

Let x^n is the leading term of polynomial function with n as highest exponent present in that function.

Then following chart summarize the limits at infinity for polynomial functions.

x—>∞ n(even)	$\dpi{120} \mathbf{\lim_{x\rightarrow \infty }x^{n}= \infty }$
x—>-∞ n(even)	$\dpi{120} \mathbf{\lim_{x\rightarrow -\infty }x^{n}= \infty }$
x—>∞ n(odd)	$\dpi{120} \mathbf{\lim_{x\rightarrow \infty }x^{n}= \infty }$
x—>-∞ n(odd)	$\dpi{120} \mathbf{\lim_{x\rightarrow -\infty }x^{n}= -\infty }$

Example1 : Find the limit of given function.

$\dpi{120} {\color{Red} \lim_{x\rightarrow \infty }x^{3}-2x+7}$
Solution: This is a polynomial function with leading term x^3, so

$\dpi{120} \lim_{x\rightarrow \infty }x^{3}-2x+7 = \lim_{x\rightarrow \infty }x^{3}=\infty$

Avoid mistakes ! In no way this can be ∞-∞=0 . This is blunder. ∞-∞ is one of the indeterminate forms.

Lets have a look at different cases of similar polynomial function.

Examples with odd degree:

$\dpi{120} \mathbf{\lim_{x\rightarrow \infty }-x^{3}+2x+7=\lim_{x\rightarrow \infty }-x^{3}=-\infty }$

$\dpi{120} \mathbf{\lim_{x\rightarrow -\infty }-x^{3}+2x+7=\lim_{x\rightarrow -\infty }-x^{3}=-\left (-\infty \right )=\infty }$

$\dpi{120} \mathbf{\lim_{x\rightarrow \infty }x^{3}+2x+7=\lim_{x\rightarrow \infty }x^{3}=\infty }$

$\dpi{120} \mathbf{\lim_{x\rightarrow -\infty }x^{3}+2x+7=\lim_{x\rightarrow -\infty }x^{3}=-\infty }$

Examples with even degree:

$\dpi{120} \mathbf{\lim_{x\rightarrow \infty }x^{2}+2x+1=\lim_{x\rightarrow \infty }x^{2}=\infty }$

$\dpi{120} \mathbf{\lim_{x\rightarrow -\infty }x^{2}+2x+1=\lim_{x\rightarrow -\infty }x^{2}=\infty }$

$\dpi{120} \mathbf{\lim_{x\rightarrow \infty }-x^{2}+2x+1=\lim_{x\rightarrow \infty }-x^{2}=-\infty }$

$\dpi{120} \mathbf{\lim_{x\rightarrow -\infty }-x^{2}+2x+1=\lim_{x\rightarrow -\infty }-x^{2}=-\infty }$

Limit at infinity of Rational functions:

To find the limit of any rational function $\dpi{120} \frac{f(x)}{g(x)}$ , we just note down the degree of numerator and denominator and follow these three rules :

If degree of numerator and denominator is same then limit is just the ratio of their leading term’s coefficients.
If degree of numerator is less than degree of denominator , then limit is 0.
If degree of numerator is more than degree of denominator then limit is either ∞ or -∞ depending on the difference in their degrees.

Lets work on some examples of each type.

Example2 : Evaluate the limit

$\dpi{120} {\color{Red} \lim_{x\rightarrow \infty }\frac{3x^{3}+5x-1}{2x^{3}-2x}}$

Solution: Proper way to solve this problem is factoring out x^3 from numerator and denominator both.

$\dpi{120} \lim_{x\rightarrow \infty }\frac{3x^{3}+5x-1}{2x^{3}-2x}=\lim_{x\rightarrow \infty }\frac{x^{3}\left ( 3+\frac{5}{x^{2}}-\frac{1}{x^{3}} \right )}{x^{3}\left ( 2-\frac{2}{x^{2}} \right )}$

$\dpi{120} =\lim_{x\rightarrow \infty }\frac{\left ( 3+\frac{5}{x^{2}}-\frac{1}{x^{3}} \right )}{\left ( 2-\frac{2}{x^{2}} \right )}$

$\dpi{120} =\frac{3+\frac{5}{\infty }-\frac{1}{\infty }}{2-\frac{2}{\infty }}=\frac{3+0-0}{2-0}=\frac{3}{2}$

Short way : As the degree of both numerator and denominator is same. So we just take the ratio of leading terms.

$\dpi{120} \lim_{x\rightarrow \infty }\frac{3x^{3}+5x-1}{2x^{3}-2x}=\lim_{x\rightarrow \infty }\frac{3x^{3}}{2x^{3}}=\lim_{x\rightarrow \infty }\frac{3}{2}=\frac{3}{2}$

Example3: Find the limit

$\dpi{120} {\color{Red} \lim_{x\rightarrow \infty }\frac{3x-1}{x^{2}-x+5}}$

Solution: Factor out x from numerator and denominator

$\dpi{120} \lim_{x\rightarrow \infty }\frac{3x-1}{x^{2}-x+5}=\lim_{x\rightarrow \infty }\frac{x\left ( 3-\frac{1}{x} \right )}{x\left ( x-1+\frac{5}{x} \right )}$

$\dpi{120} =\lim_{x\rightarrow \infty }\frac{\left ( 3-\frac{1}{x} \right )}{\left ( x-1+\frac{5}{x} \right )}$

$\dpi{120} =\frac{3-\frac{1}{\infty }}{\infty -1+\frac{5}{\infty }}=\frac{3-0}{\infty -1+0}= \frac{3}{\infty }=0$

Short way: Degree of numerator is 1 which is less than degree of denominator(2) so limit would be just 0.

Example4: Find the limit

$\dpi{120} {\color{Red} \lim_{x\rightarrow -\infty }\frac{2x^{3}-x+3}{x+5}}$
Solution: Factor out x from numerator and denominator.

$\dpi{120} \lim_{x\rightarrow -\infty }\frac{2x^{3}-x+3}{x+5}= \lim_{x\rightarrow -\infty }\frac{x\left ( 2x^{2}-1 +\frac{3}{x}\right )}{x\left ( 1+\frac{5}{x} \right )}$

$\dpi{120} = \lim_{x\rightarrow -\infty }\frac{\left ( 2x^{2}-1 +\frac{3}{x}\right )}{\left ( 1+\frac{5}{x} \right )}$

$\dpi{120} =\frac{2(-\infty )^{2}-1+\frac{3}{-\infty }}{1+\frac{5}{-\infty }}=\frac{\infty -1-0}{1+0}=\frac{\infty }{1}=\infty$

Short way: Since the difference between degrees of numerator and denominator is 3-1 =2 (even)

And we know that even power(exponent) always give a positive result even when x is approaching to any negative value.

So $\dpi{120} \lim_{x\rightarrow -\infty }\frac{2x^{3}-x+3}{x+5}= \lim_{x\rightarrow -\infty }2x^{2}=\infty$

When difference of degrees is an odd number with x approaching to -∞ then answer would be -∞

Limits at infinity for square root functions:

For square root functions, we use rationalizing.

Example5: Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow \infty }\sqrt{x^{2}+x+1}-\sqrt{x^{2}+1}}$

Solution: Here the expression assumes the form ∞-∞- so first reduce it to form $\dpi{120} \frac{f(x)}{g(x)}$ using rationalizing.

$\dpi{120} \lim_{x\rightarrow \infty }\frac{\sqrt{x^{2}+x+1}-\sqrt{x^{2}+1}}{1}*\frac{\sqrt{x^{2}+x+1}+\sqrt{x^{2}+1}}{\sqrt{x^{2}+x+1}+\sqrt{x^{2}+1}}$ [rationalizing numerator]

$\dpi{120} \lim_{x\rightarrow \infty }\frac{\left ( \sqrt{x^{2}+x+1} \right )^{2}-\left ( \sqrt{x^{2}+1} \right )^{2}}{\sqrt{x^{2}+x+1}+\sqrt{x^{2}+1}}$

$\dpi{120} \lim_{x\rightarrow \infty }\frac{\left (x^{2}+x+1 \right )-(x^{2}+1)}{\sqrt{x^{2}+x+1}+\sqrt{x^{2}+1}}$

$\dpi{120} \lim_{x\rightarrow \infty }\frac{x^{2}+x+1 -x^{2}-1}{\sqrt{x^{2}+x+1}+\sqrt{x^{2}+1}}=\lim_{x\rightarrow \infty }\frac{x}{\sqrt{x^{2}+x+1}+\sqrt{x^{2}+1}}$

$\dpi{120} =\lim_{x\rightarrow \infty }\frac{x}{\sqrt{x^{2}\left (1+\frac{1}{x}+\frac{1}{x^{2}} \right )}+\sqrt{x^{2}\left (1+\frac{1}{x^{2}} \right )}}$

$\dpi{120} =\lim_{x\rightarrow \infty }\frac{x}{x\left (\sqrt{\left (1+\frac{1}{x}+\frac{1}{x^{2}} \right )}+\sqrt{\left (1+\frac{1}{x^{2}} \right )} \right )}$

$\dpi{120} =\lim_{x\rightarrow \infty }\frac{1}{\left (\sqrt{1+\frac{1}{x}+\frac{1}{x^{2}} }+\sqrt{1+\frac{1}{x^{2}}} \right )}$

$\dpi{120} =\frac{1}{\sqrt{1+0+0}+\sqrt{1+0}}=\frac{1}{2}$

Example6 : Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow -\infty }\sqrt{x^{2}-x+1}+x}$

Solution:

$\dpi{120} \lim_{x\rightarrow -\infty }\frac{\sqrt{x^{2}-x+1}+x}{1}*\frac{\sqrt{x^{2}-x+1}-x}{\sqrt{x^{2}-x+1}-x}$

$\dpi{120} \lim_{x\rightarrow -\infty }\frac{\left ( \sqrt{x^{2}-x+1} \right )^{2}-\left ( x \right )^{2}}{\sqrt{x^{2}-x+1}-x}$

$\dpi{120} \lim_{x\rightarrow -\infty }\frac{ x^{2}-x+1 -x ^{2}}{\sqrt{x^{2}\left ( 1-\frac{1}{x}+\frac{1}{x^{2}} \right )}-x}$

$\dpi{120} \lim_{x\rightarrow -\infty }\frac{ -x+1 }{\left | x \right |\sqrt{\left ( 1-\frac{1}{x}+\frac{1}{x^{2}} \right )}-x}$

$\dpi{120} \lim_{x\rightarrow -\infty }\frac{ -x+1 }{\left | x \right |\left (\sqrt{\left ( 1-\frac{1}{x}+\frac{1}{x^{2}} \right )}-\frac{x}{\left | x \right |} \right )}$

Note: As radicals are not defined for negative numbers so $\dpi{120} \mathbf{x^{2}=\left | x \right |}$

Also |x| = -x for all x < 0 therefore $\dpi{120} \mathbf{\frac{x}{\left | x \right |}=-1}$ for x < 0

$\dpi{120} \lim_{x\rightarrow -\infty }\frac{ \frac{-x}{\left | x \right |}+\frac{1}{\left | x \right |} }{\left (\sqrt{\left ( 1-\frac{1}{x}+\frac{1}{x^{2}} \right )}-\frac{x}{\left | x \right |} \right )}$

$\dpi{120} \frac{-(-1)+0}{\sqrt{1-0-0}-(-1)}=\frac{1}{1+1}=\frac{1}{2}$

Practice problems:

Evaluate the limit of following expressions.

$\dpi{120} \lim_{x\rightarrow -\infty }2x^{5}-3x^{2}-x+5$

$\dpi{120} lim_{x\rightarrow \infty }x^{4}-3x+1$

$\dpi{120} \lim_{x\rightarrow \infty }\frac{5x-x^{2}-4x^{3}}{5x^{3}-2x}$

$\dpi{120} \lim_{x\rightarrow \infty }\frac{3x-x^{2}}{x^{3}+2x}$

$\dpi{120} \lim_{x\rightarrow \infty }\frac{2x-x^{3}}{1-2x}$

$\dpi{120} \lim_{x\rightarrow -\infty }x+\sqrt{x^{2}+2x}$

Answers:

-∞
∞
-4/5
0
∞
-1

Algebraic Limits and its Properties

September 18, 2019/by admin_ct

What is limit ?

In simple words, limit is the output(y value) that a function approaches as the input(x value) approaches some value.

Limit of a function f(x) is usually written as

$\dpi{150} \mathbf{\lim_{x\rightarrow c}f(x) = L}$

And is read as limit of f(x) as x approaches to c, is L

Which can be easily understood as when x approaches to c, y approaches to L.

There are many techniques for evaluating limits. But before that we should get familiar with properties of limits.

Properties of limits

Let $\dpi{150} \mathbf{\lim_{x\rightarrow a}f(x)}$ and $\dpi{150} \mathbf{\lim_{x\rightarrow c}f(x) }$ exist and let c be a constant.

$\dpi{150} \mathbf{\lim_{x\rightarrow a}\left [ f(x)\pm g(x) \right ]= \lim_{x\rightarrow a}f(x)\pm \lim_{x\rightarrow a}g(x))}$

$\dpi{150} \mathbf{\lim_{x\rightarrow a} cf(x)= c\lim_{x\rightarrow a}f(x)}$

$\dpi{150} \mathbf{\lim_{x\rightarrow a}\left [ f(x)g(x) \right ]=\lim_{x\rightarrow a}f(x)*\lim_{x\rightarrow a}g(x)}$

$\dpi{150} \mathbf{\lim_{x\rightarrow a}\left [ \frac{f(x)}{g(x)} \right ]= \frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)} if \lim_{x\rightarrow a}g(x)\neq 0}$

$\dpi{150} \mathbf{\lim_{x\rightarrow a}\left [ f(x) \right ]^{n}=\left [ \lim_{x\rightarrow a}f(x) \right ]^{n}}$ where n is a positive integer.

$\dpi{150} \mathbf{\lim_{x\rightarrow a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x\rightarrow a}f(x)}}$

$\dpi{150} \mathbf{\lim_{x\rightarrow a}c=c}$ [limit of constant is constant always]

Different methods to evaluate the algebraic limits are:

Limit by direct substitution
Limit by factorization
Limit by Rationalizing
By using some standard limits

Lets discuss them one by one.

1) Limit by direct substitution:

There are many well behaved functions such as polynomials , rational functions with non zero denominators for which we can use direct substitution to evaluate limit.

Example1: Evaluate the limit.

$\dpi{120} {\color{Red} \lim_{x\rightarrow 1}3x^{2}+4x+5}$

Solution: As this is a polynomial function, we just substitute x as 1 and solve.

$\dpi{120} \lim_{x\rightarrow 1}3x^{2}+4x+5 = 3(1)^{2}+4(1)+5 =12$

Example2 : Evaluate the limit.

$\dpi{120} {\color{Red} \lim_{x\rightarrow 2}\frac{x^{2}-4}{x+3}}$

Solution: This is a rational function with non zero denominator.

So, $\dpi{120} \lim_{x\rightarrow 2}\frac{x^{2}-4}{x+3}=\frac{2^{2}-4}{2+3}= \frac{0}{5}=0$

2) Limit by factorization :

For this method, one should be familiar with all factoring techniques. If after substituting limit x->a into given rational function we get form 0/0 and (x-a) is a common factor of numerator and denominator both, then we cancel out the common factor and substitute limit in simplified form to get final answer.

Example3: Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow 2}\frac{x^{2}-5x+6}{x^{2}-4}}$

Solution: When we plugin limit x=2 in this function, we get 0/0 . That means (x-2) is a common factor here so we try to factor numerator and denominator .

$\dpi{120} \lim_{x\rightarrow 2}\frac{x^{2}-5x+6}{x^{2}-4}= \lim_{x\rightarrow 2}\frac{(x-2)(x-3)}{(x-2)(x+2)} =\lim_{x\rightarrow 2}\frac{x-3}{x+2}=\frac{2-3}{2+2}=\frac{-1}{4}$

Example4: Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow 1}\frac{2}{1-x^{2}}+\frac{1}{x-1}}$

Solution: When we plugin limit x=1, it takes the form ∞+∞ . So we need to simplify it further to express it as 0/0.

$\dpi{120} \lim_{x\rightarrow 1}\frac{2}{1-x^{2}}+\frac{1}{x-1}=\lim_{x\rightarrow 1}\left [ \frac{2-(1+x)}{1-x^{2}} \right ]$

$\dpi{120} \Rightarrow \lim_{x\rightarrow 1}\frac{1-x}{1-x^{2}}=\lim_{x\rightarrow 1}\frac{1-x}{(1-x)(1+x)}$

$\dpi{120} \Rightarrow \lim_{x\rightarrow 1}\frac{1}{1+x}=\frac{1}{1+1}=\frac{1}{2}$

3) Limit by Rationalizing :

This method is used when either the numerator or denominator or both involve square roots and substituting the limit , expression take the form 0/0. Lets look at some examples.

Example5 : Evaluate the limit .

$\dpi{120} {\color{Red} \lim_{x\rightarrow 0}\frac{\sqrt{2+x}-\sqrt{2}}{x}}$

Solution: Here we rationalize the numerator.

$\dpi{120} \lim_{x\rightarrow 0}\frac{\sqrt{2+x}-\sqrt{2}}{x}=\lim_{x\rightarrow 0}\frac{\sqrt{2+x}-\sqrt{2}}{x}*\frac{\sqrt{2+x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{2}}$

$\dpi{120} =\lim_{x\rightarrow 0}\frac{\left (\sqrt{2+x}-\sqrt{2} \right )\left ( \sqrt{2+x}+\sqrt{2} \right )}{x\left (\sqrt{2+x}+\sqrt{2} \right )}$

$\dpi{120} =\lim_{x\rightarrow 0}\frac{\left (\sqrt{2+x} \right )^{2}-\left ( \sqrt{2} \right )^{2}}{x\left ( \sqrt{2+x}+\sqrt{2} \right )}$

$\dpi{120} =\lim_{x\rightarrow 0}\frac{2+x-2}{x\left ( \sqrt{2+x}+\sqrt{2} \right )}=\lim_{x\rightarrow 0}\frac{x}{x\left ( \sqrt{2+x}+\sqrt{2} \right )}$

$\dpi{120} =\lim_{x\rightarrow 0}\frac{1}{\sqrt{2+x}+\sqrt{2}}=\frac{1}{\sqrt{2}+\sqrt{2}}=\frac{1}{2\sqrt{2}}$

Example6: Find the limit

$\dpi{120} {\color{Red} \lim_ {x\rightarrow 7}\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}}$

Solution: Here we rationalize both numerator and denominator.

$\dpi{120} \lim_ {x\rightarrow 7}\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}=\lim_ {x\rightarrow 7}\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}*\frac{1+\sqrt{8-x}}{1+\sqrt{8-x}}*\frac{4+\sqrt{9+x}}{4+\sqrt{9+x}}$

$\dpi{120} =\lim_{x\rightarrow 7}\frac{\ 4^{2}-\left ( \sqrt{9+x} \right )^{2}\left ( 1+\sqrt{8-x} \right )}{1^{2}-\left ( \sqrt{8-x} \right )^{2}\left ( 4+\sqrt{9+x} \right )}$

$\dpi{120} =\lim_{x\rightarrow 7}\frac{(16-9-x)\left ( 1+\sqrt{8-x} \right )}{(1-8+x)\left ( 4+\sqrt{9+x} \right )}$

$\dpi{120} =\lim_{x\rightarrow 7}\frac{(7-x)\left ( 1+\sqrt{8-x} \right )}{(-7+x)\left ( 4+\sqrt{9+x} \right )}=\lim_{x\rightarrow 7}\frac{-(-7+x)\left ( 1+\sqrt{8-x} \right )}{(-7+x)\left ( 4+\sqrt{9+x} \right )}$

$\dpi{120} =\lim_{x\rightarrow 7}\frac{-\left (1+\sqrt{8-x} \right )}{4+\sqrt{9+x}}= \frac{-\left (1+\sqrt{1} \right )}{4+\sqrt{16}}$

$\dpi{120} =\frac{-2}{4+4} =\frac{-1}{4}$

4) Using standard limits:

Here we use this formula:

$\dpi{150} \mathbf{\lim_{x\rightarrow a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}}$

Example7: Evaluate limit

$\dpi{120} {\color{Red} \lim_{x\rightarrow 2}\frac{x^{10}-1024}{x-2}}$

Solution: When we plugin limit x=2, this expression takes the form 0/0 so we use standard formula.

$\dpi{120} lim_{x\rightarrow 2}\frac{x^{10}-1024}{x-2}= \lim_{x\rightarrow 2}\frac{x^{10}-2^{10}}{x-2}$

$\dpi{120} = 10*2^{10-1}=10*2^{9}= 5120$

Example8: If $\dpi{120} {\color{Red} \lim_{x\rightarrow a}\frac{x^{3}-a^{3}}{x-a}=\lim_{x\rightarrow a}\frac{x^{4}-1}{x-1}}$ then find all possible values of a.