Limits Definition And Complete Solution


Limit / Wednesday, August 15th, 2018
(Last Updated On: August 15, 2018)

Limit

Let us first understand the limits definition, and then we learn all the limits formulas. After that we will try to solve various limits problems which appear in various exams. I will try to solve as much as I can here, if anything left out you can comment, I will try to solve that in next update. So let’s start:

Limits Definition

The limit of a function f(x) for a given value ‘a’ of the independent variable x is the constant ‘l’ for which the function f(x) can be made as small as we please by making x to approach sufficiently nearer to its assigned value a (denoted as x → a). It can be expressed as

\[\underset{x\to a}{\mathop{\lim }}\,f(x)=l\]

That is, a function f(x) is said to have a limit ‘l’ as x approaches ‘a’, if for any positive number ϵ, there exists a number δ depending on ϵ can be found out, such that |f(x) – l| < ϵ for all values of x for which 0<|x – a|<δ
This definition is called Cauchy’s definition of a limit of a function.

 Question 01

Define the symbol, \[\underset{x\to a+0}{\mathop{\lim }}\,f(x)={{l}_{1}}\]

The symbol means given any pre-assigned positive quantity ϵ, however small, we can determine a positive quantity δ, such that |f(x) – l1|<ϵ whenever 0<x-a<=δ i.e., a<x<=a+δ.

The given symbol sometimes denoted by the symbol f(a+0). It is also called Right-hand limit of f(x) as x tends to a.

 Question 02

Define the symbol, \[\underset{x\to a-0}{\mathop{\lim }}\,f(x)={{l}_{2}}\]

The symbol means given any pre-assigned positive quantity ϵ, however small, we can determine a positive quantity δ, such that | l2 – f(x)|-<ϵ whenever 0<a-x<=δ i.e., a-δ<=x<a.

The given symbol sometimes denoted by the symbol f(a-0). It is also called Left-hand limit of f(x) as x tends to a.

 Question 03

Distinguish between, \[f(a),and\underset{x\to a}{\mathop{\lim }}\,f(x)\]

\[\underset{x\to a}{\mathop{\lim }}\,f(x)\] \[f(a)\]
Is a statement about the value of f(x) when x has any value arbitrarily near to ‘a’, except ‘a’, in this case we do not know what happens when x is put equal to ‘a’. It stands for the value of f(x) when x is exactly equal to ‘a’, obtained either by the definition of the function at ‘a’, or else by the substitution of ‘a’ for x in the expression f(x), when it exists.

Fundamental theorems of limit

Let,c be a finite constant, \[\underset{x\to a}{\mathop{\lim }}\,f(x)=l,and\underset{x\to a}{\mathop{\lim }}\,\phi (x)=m\]
Then,
\[(i)\underset{x\to a}{\mathop{\lim }}\,cf(x)=c.\underset{x\to a}{\mathop{\lim }}\,f(x)=cl\]
\[(ii)\underset{x\to a}{\mathop{\lim }}\,[f(x)+\phi (x)]=\underset{x\to a}{\mathop{\lim }}\,f(x)+\underset{x\to a}{\mathop{\lim }}\,\phi (x)=l+m\]
\[(iii)\underset{x\to a}{\mathop{\lim }}\,[f(x)-\phi (x)]=\underset{x\to a}{\mathop{\lim }}\,f(x)-\underset{x\to a}{\mathop{\lim }}\,\phi (x)=l-m\]
\[(iv)\underset{x\to a}{\mathop{\lim }}\,[f(x).\phi (x)]=\underset{x\to a}{\mathop{\lim }}\,f(x).\underset{x\to a}{\mathop{\lim }}\,\phi (x)=l.m\]
\[(v)\underset{x\to a}{\mathop{\lim }}\,\left[ \frac{\underset{x\to a}{\mathop{\lim }}\,f(x)}{\underset{x\to a}{\mathop{\lim }}\,\phi (x)} \right]=\frac{\underset{x\to a}{\mathop{\lim }}\,f(x)}{\underset{x\to a}{\mathop{\lim }}\,\phi (x)}=\frac{l}{m},[m\ne 0]\]
\[(vi)\underset{x\to a}{\mathop{\lim }}\,F\{f(x)\}=F\{\underset{x\to a}{\mathop{\lim }}\,f(x)\}=F(l)\]
Where F(u) is a function of u which is continuous for u=l.

Some important limits

\[(i)\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1\]
\[(ii)\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}\]
\[(iii)\underset{x\to \infty }{\mathop{\lim }}\,f(x)=\underset{z\to 0}{\mathop{\lim }}\,f\left( \frac{1}{z} \right),\left[ \because x=\frac{1}{z} \right]\]
\[(iv)\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-1}{x}=1.\]
\[(v)\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}(1+x)}{x}=1.\]
\[(vi)\underset{x\to 0}{\mathop{\lim }}\,{{(1+x)}^{\frac{1}{x}}}=e\]
\[(vii)\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{x} \right)}^{x}}=e\]
\[(viii)\underset{x\to 0}{\mathop{\lim }}\,\frac{{{a}^{x}}-1}{x}={{\log }_{e}}a,(a>0)\]
\[(ix)\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}(1+x)-1}{x}=n.\]

L. Hospital’s theorem

Let f(x) and φ(x) be the functions of x which can be expanded by Taylor’s theorem and also let f(a) = 0 and φ(a) = 0, then,\[\underset{x\to a}{\mathop{\lim }}\,\frac{f(x)}{\phi (x)}=\frac{f'(x)}{\phi ‘(x)}\]provided the later limit exists.

 Question 04

Evaluate:
\[(i)\underset{x\to \frac{a}{2}}{\mathop{\lim }}\,\frac{8{{x}^{2}}-10ax+3{{a}^{2}}}{4{{x}^{2}}+4ax-3{{a}^{2}}}\]
\[(ii)\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+2{{x}^{2}}}-\sqrt{1-2{{x}^{2}}}}{{{x}^{2}}}\]
\[(iii)\underset{x\to 0}{\mathop{\lim }}\,\frac{{{(1+x)}^{\frac{1}{2}}}-1}{{{(1+x)}^{\frac{1}{3}}}-1}\]

Solution:
\[(i)\underset{x\to \frac{a}{2}}{\mathop{\lim }}\,\frac{8{{x}^{2}}-10ax+3{{a}^{2}}}{4{{x}^{2}}+4ax-3{{a}^{2}}}\]
If we put a/2 in x in the denominator we get,
\[4.\frac{{{a}^{2}}}{4}+4a.\frac{a}{2}-3{{a}^{2}}=0\]
So, we have to factorize both numerator and denominator.
\[\therefore 8{{x}^{2}}-10ax+3{{a}^{2}}=8{{x}^{2}}-(6+4)ax+3{{a}^{2}}\]
\[=8{{x}^{2}}-6ax-4ax+3{{a}^{2}}\]
\[=2x(4x-3a)-a(4x-3a)\]
\[=(4x-3a)(2x-a)\]
\[\therefore 4{{x}^{2}}+4ax-3{{a}^{2}}=4{{x}^{2}}+(6-2)ax-3{{a}^{2}}\]
\[=4{{x}^{2}}+6ax-2ax-3{{a}^{2}}\]
\[=2x(2x+3a)-a(2x+3a)\]
\[=(2x+3a)(2x-a)\]
Now we get,
\[\underset{x\to \frac{a}{2}}{\mathop{\lim }}\,\frac{8{{x}^{2}}-10ax+3{{a}^{2}}}{4{{x}^{2}}+4ax-3{{a}^{2}}}\]
\[=\underset{x\to \frac{a}{2}}{\mathop{\lim }}\,\frac{(4x-3a)(2x-a)}{(2x+3a)(2x-a)}\]
\[=\underset{x\to \frac{a}{2}}{\mathop{\lim }}\,\frac{(4x-3a)}{(2x+3a)}\]
\[=\frac{(4.\frac{a}{2}-3a)}{(2.\frac{a}{2}+3a)}=\frac{-a}{4a}=\frac{-1}{4}\]

Solution:

\[(ii)\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+2{{x}^{2}}}-\sqrt{1-2{{x}^{2}}}}{{{x}^{2}}}\]
\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{(\sqrt{1+2{{x}^{2}}}-\sqrt{1-2{{x}^{2}}})(\sqrt{1+2{{x}^{2}}}+\sqrt{1-2{{x}^{2}}})}{{{x}^{2}}(\sqrt{1+2{{x}^{2}}}+\sqrt{1-2{{x}^{2}}})}\]
\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\left( \sqrt{1+2{{x}^{2}}} \right)}^{2}}-{{\left( \sqrt{1-2{{x}^{2}}} \right)}^{2}}}{{{x}^{2}}(\sqrt{1+2{{x}^{2}}}+\sqrt{1-2{{x}^{2}}})}\]
\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1+2{{x}^{2}}-1+2{{x}^{2}}}{{{x}^{2}}(\sqrt{1+2{{x}^{2}}}+\sqrt{1-2{{x}^{2}}})}\]
\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{4{{x}^{2}}}{{{x}^{2}}(\sqrt{1+2{{x}^{2}}}+\sqrt{1-2{{x}^{2}}})}\]
\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{4}{(\sqrt{1+2{{x}^{2}}}+\sqrt{1-2{{x}^{2}}})}\]
\[=\frac{4}{(\sqrt{1+2.0}+\sqrt{1-2.0})}\]
\[=\frac{4}{2}=2\]

Solution:
\[(iii)\underset{x\to 0}{\mathop{\lim }}\,\frac{{{(1+x)}^{\frac{1}{2}}}-1}{{{(1+x)}^{\frac{1}{3}}}-1}\]
Let z = 1+x, then z→1, when x→0
\[\therefore \underset{x\to 0}{\mathop{\lim }}\,\frac{{{(1+x)}^{\frac{1}{2}}}-1}{{{(1+x)}^{\frac{1}{3}}}-1}=\underset{z\to 1}{\mathop{\lim }}\,\frac{{{z}^{\frac{1}{2}}}-1}{{{z}^{\frac{1}{3}}}-1}\]
\[=\underset{z\to 1}{\mathop{\lim }}\,\frac{\frac{{{z}^{\frac{1}{2}}}-1}{z-1}}{\frac{{{z}^{\frac{1}{3}}}-1}{z-1}}\]
\[=\frac{\underset{z\to 1}{\mathop{\lim }}\,\frac{{{z}^{\frac{1}{2}}}-1}{z-1}}{\underset{z\to 1}{\mathop{\lim }}\,\frac{{{z}^{\frac{1}{3}}}-1}{z-1}}\]
\[=\frac{\frac{1}{2}{{.1}^{\frac{1}{2}-1}}}{\frac{1}{3}{{.1}^{\frac{1}{3}-1}}},\left[ \because \underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}} \right]\]
\[=\frac{\frac{1}{2}}{\frac{1}{3}}=\frac{3}{2}\]

 Question 05

Prove that,
\[(i)\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}=\frac{1}{3}{{x}^{\frac{-2}{3}}}\]
\[(ii)\underset{x\to 3}{\mathop{\lim }}\,\frac{\sqrt{x-3}+\sqrt{x}-\sqrt{3}}{\sqrt{{{x}^{2}}-9}}=\frac{1}{\sqrt{6}}\]
Solution:
\[(i)\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}=\frac{1}{3}{{x}^{\frac{-2}{3}}}\]
Let x+h=z, then z→x as h→0 and h=z-x
\[\therefore \underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}=\underset{z\to x}{\mathop{\lim }}\,\frac{\sqrt[3]{z}-\sqrt[3]{x}}{z-x}\]
\[=\underset{z\to x}{\mathop{\lim }}\,\frac{{{z}^{\frac{1}{3}}}-{{x}^{\frac{1}{3}}}}{z-x}=\frac{1}{3}.{{x}^{\frac{1}{3}-1}},\left[ \because \underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}} \right]\]
\[=\frac{1}{3}{{x}^{-\frac{2}{3}}}(\Pr oved)\]

Solution:
\[(ii)\underset{x\to 3}{\mathop{\lim }}\,\frac{\sqrt{x-3}+\sqrt{x}-\sqrt{3}}{\sqrt{{{x}^{2}}-9}}=\frac{1}{\sqrt{6}}\]
\[L.H.S.=\underset{x\to 3}{\mathop{\lim }}\,\frac{\sqrt{x-3}+\sqrt{x}-\sqrt{3}}{\sqrt{{{x}^{2}}-9}}\]
\[=\underset{x\to 3}{\mathop{\lim }}\,\frac{\sqrt{x-3}}{\sqrt{{{x}^{2}}-9}}+\underset{x\to 3}{\mathop{\lim }}\,\frac{\sqrt{x}-\sqrt{3}}{\sqrt{{{x}^{2}}-9}}\]
\[=\underset{x\to 3}{\mathop{\lim }}\,\frac{\sqrt{x-3}}{\sqrt{x-3}\sqrt{x+3}}+\underset{x\to 3}{\mathop{\lim }}\,\frac{\left[ (\sqrt{x}-\sqrt{3})(\sqrt{x}+\sqrt{3}) \right]\sqrt{{{x}^{2}}-9}}{\sqrt{{{x}^{2}}-9}\sqrt{{{x}^{2}}-9}(\sqrt{x}+\sqrt{3})}\]
\[=\underset{x\to 3}{\mathop{\lim }}\,\frac{1}{\sqrt{x+3}}+\underset{x\to 3}{\mathop{\lim }}\,\frac{(x-3)\sqrt{{{x}^{2}}-9}}{({{x}^{2}}-9)(\sqrt{x}+\sqrt{3})}\]
\[=\underset{x\to 3}{\mathop{\lim }}\,\frac{1}{\sqrt{x+3}}+\underset{x\to 3}{\mathop{\lim }}\,\frac{(x-3)\sqrt{{{x}^{2}}-9}}{(x-3)(x+3)(\sqrt{x}+\sqrt{3})}\]
\[=\frac{1}{\sqrt{3+3}}+\underset{x\to 3}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}-9}}{(x+3)(\sqrt{x}+\sqrt{3})}\]
\[=\frac{1}{\sqrt{3+3}}+\frac{\sqrt{9-9}}{(3+3)(\sqrt{3}+\sqrt{3})}\]
\[=\frac{1}{\sqrt{6}}+0=\frac{1}{\sqrt{6}}=R.H.S.\]

 Question 06

Evaluate
\[(i)\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}-\tan 2x}{\tan x}\]
\[(ii)\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos 5x-\cos 7x}{\cos x-\cos 5x}\]
\[(iii)\underset{x\to \pi }{\mathop{\lim }}\,\frac{1+\cos x}{\pi -x}\]

Solution:
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}-\tan 2x}{\tan x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{x\left( x-\frac{\tan 2x}{x} \right)}{\tan x}\]
\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( x-\frac{\tan 2x}{x} \right)}{\frac{\tan x}{x}},[\because x\to 0\Rightarrow x\ne 0]\]
\[=\frac{\underset{x\to 0}{\mathop{\lim }}\,x-\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin 2x}{2x}\times 2\times \underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\cos 2x}}{\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\cos x}}\]
\[=\frac{0-\underset{u\to 0}{\mathop{\lim }}\,\frac{\sin u}{u}\times 2\times 1}{1\times 1},[let,u=2x,\therefore x\to 0\Rightarrow u\to 0]\]
\[=-1\times 2=-2\]

Solution:
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos 5x-\cos 7x}{\cos x-\cos 5x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin 6x\sin x}{2\sin 3x\sin 2x}\]
\[\left[ \because \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2} \right]\]
\[=\frac{\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin 6x}{6x}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}}{\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin 3x}{3x}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin 2x}{2x}}\]
\[=\frac{1\times 1}{1\times 1}=1\]

Solution:

Let, z=π-x, then z→0 as x→πand x=π-z
\[\underset{x\to \pi }{\mathop{\lim }}\,\frac{1+\cos x}{\pi -x}=\underset{z\to 0}{\mathop{\lim }}\,\frac{1+\cos (\pi -z)}{z}\]
\[=\underset{z\to 0}{\mathop{\lim }}\,\frac{1-\cos z}{z}\]
\[=\underset{z\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\frac{z}{2}}{\frac{{{z}^{2}}}{4}\times \frac{4}{z}}\]
\[=\underset{z\to 0}{\mathop{\lim }}\,\frac{2z}{4}\times {{\left[ \underset{z\to 0}{\mathop{\lim }}\,\frac{\sin \frac{z}{2}}{\frac{z}{2}} \right]}^{2}}\]
\[=\frac{2\times 0}{4}\times {{(1)}^{2}}=0\]

 

 Question 07

Prove that,
\[(i)\underset{x\to a}{\mathop{\lim }}\,\frac{\sin x-\sin a}{x-a}=\cos a\]
\[(ii)\underset{x\to y}{\mathop{\lim }}\,\frac{{{\cos }^{2}}x-{{\cos }^{2}}y}{{{x}^{2}}-{{y}^{2}}}=-\frac{\sin 2y}{2y}\]
\[(iii)\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan 2x-x}{3x-\sin x}=\frac{1}{2}\]
Solution:
\[L.H.S.=\underset{x\to a}{\mathop{\lim }}\,\frac{\sin x-\sin a}{x-a}\]
\[=\underset{x\to a}{\mathop{\lim }}\,\frac{2\cos \frac{x+a}{2}\sin \frac{x-a}{2}}{\frac{x-a}{2}\times 2}\]
\[=\underset{x\to a}{\mathop{\lim }}\,\cos \frac{x+a}{2}.\underset{x\to a}{\mathop{\lim }}\,\frac{\sin \frac{x-a}{2}}{\frac{x-a}{2}}\]
Let z=(x-a)/2, then z→0 as x→a
\[=\cos \frac{a+a}{2}.\underset{z\to 0}{\mathop{\lim }}\,\frac{\sin z}{z}\]
\[=\cos a.1,\left[ \because \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1 \right]\]
\[=\cos a=R.H.S.\]

Solution:
\[L.H.S=\underset{x\to y}{\mathop{\lim }}\,\frac{{{\cos }^{2}}x-{{\cos }^{2}}y}{{{x}^{2}}-{{y}^{2}}}\]
\[=\underset{x\to y}{\mathop{\lim }}\,\frac{(\cos x+\cos y)(\cos x-\cos y)}{{{x}^{2}}-{{y}^{2}}}\]
\[=\underset{x\to y}{\mathop{\lim }}\,(\cos x+\cos y)\times \underset{x\to y}{\mathop{\lim }}\,\frac{2\sin \frac{x+y}{2}\sin \frac{y-x}{2}}{(x+y)(x-y)}\]
\[=\underset{x\to y}{\mathop{\lim }}\,(\cos x+\cos y)\times 2\times \underset{x\to y}{\mathop{\lim }}\,\frac{\sin \frac{x+y}{2}}{(x+y)}\times \underset{x\to y}{\mathop{\lim }}\,\frac{-\sin \frac{x-y}{2}}{\frac{(x-y)}{2}\times 2}\]
Let z=(x-y)/2, then z→0 as x→y
\[=-2(\cos y+\cos y)\times \frac{\sin \frac{y+y}{2}}{y+y}\times \frac{1}{2}\times \underset{z\to 0}{\mathop{\lim }}\,\frac{\sin z}{z}\]
\[=-2\times \frac{1}{2}\times 2\cos y\times \frac{\sin y}{2y}=-\frac{2\sin y\cos y}{2y}\]
\[=-\frac{\sin 2y}{2y}=R.H.S.\]

Solution:
\[L.H.S.=\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan 2x-x}{3x-\sin x}\]
\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{x\left( \frac{\tan 2x}{x}-1 \right)}{x\left( 3-\frac{\sin x}{x} \right)}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{\sin 2x}{x\cos 2x}-1 \right)}{\left( 3-\frac{\sin x}{x} \right)}\]
\[=\frac{\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin 2x}{2x}\times 2\times \underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\cos 2x}-\underset{x\to 0}{\mathop{\lim }}\,1}{\underset{x\to 0}{\mathop{\lim }}\,3-\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}}\]
\[=\frac{1\times 2\times \frac{1}{1}-1}{3-1}=\frac{1}{2}=R.H.S.\]

 Question 08

Evaluate:
\[(i)\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+…+{{n}^{3}}}{{{n}^{4}}}\]
\[(ii)\underset{x\to \infty }{\mathop{\lim }}\,{{e}^{\frac{1}{{{x}^{2}}}+2}}\]
\[(iii)\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sqrt{1+{{n}^{2}}}-\sqrt{1+n}}{\sqrt{1+{{n}^{3}}}-\sqrt{1+n}}\]
Solution:
\[\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+…+{{n}^{3}}}{{{n}^{4}}}\]
\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{\left[ \frac{n(n+1)}{2} \right]}^{2}}}{{{n}^{4}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}}{{{n}^{4}}}\]
\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{n}^{2}}+2n+1}{4{{n}^{2}}}\]
\[=\frac{1}{4}\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\frac{2}{n}+\frac{1}{{{n}^{2}}} \right]\]
\[=\frac{1}{4}(1+0+0)=\frac{1}{4},\left[ \because \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}=0\And \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{{{n}^{2}}}=0 \right]\]

Solution:
Let, z=1/x. then z→0 as x→∞
\[\underset{x\to \infty }{\mathop{\lim }}\,{{e}^{\frac{1}{{{x}^{2}}}+2}}=\underset{z\to 0}{\mathop{\lim }}\,{{e}^{{{z}^{2}}+2}}\]
\[={{e}^{0+2}}={{e}^{2}}\]

Solution:
\[\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sqrt{1+{{n}^{2}}}-\sqrt{1+n}}{\sqrt{1+{{n}^{3}}}-\sqrt{1+n}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sqrt{{{n}^{2}}\left( \frac{1}{{{n}^{2}}}+1 \right)}-\sqrt{n\left( \frac{1}{n}+1 \right)}}{\sqrt{{{n}^{3}}\left( \frac{1}{{{n}^{3}}}+1 \right)}-\sqrt{n\left( \frac{1}{n}+1 \right)}}\]
\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n.\sqrt{\left( \frac{1}{{{n}^{2}}}+1 \right)}-\sqrt{n}\sqrt{\left( \frac{1}{n}+1 \right)}}{{{n}^{\frac{3}{2}}}.\sqrt{\left( \frac{1}{{{n}^{2}}}+1 \right)}-\sqrt{n}\sqrt{\left( \frac{1}{n}+1 \right)}}\]
\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n.\left[ \sqrt{\left( \frac{1}{{{n}^{2}}}+1 \right)}-\frac{1}{\sqrt{n}}\sqrt{\left( \frac{1}{n}+1 \right)} \right]}{{{n}^{\frac{3}{2}}}.\left[ \sqrt{\left( \frac{1}{{{n}^{2}}}+1 \right)}-\frac{1}{n}\sqrt{\left( \frac{1}{n}+1 \right)} \right]}\]
\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\left[ \sqrt{\left( \frac{1}{{{n}^{2}}}+1 \right)}-\frac{1}{\sqrt{n}}\sqrt{\left( \frac{1}{n}+1 \right)} \right]}{{{n}^{\frac{1}{2}}}.\left[ \sqrt{\left( \frac{1}{{{n}^{2}}}+1 \right)}-\frac{1}{n}\sqrt{\left( \frac{1}{n}+1 \right)} \right]}\]
\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{\sqrt{n}}\times \underset{n\to \infty }{\mathop{\lim }}\,\frac{\left[ \sqrt{\left( \frac{1}{{{n}^{2}}}+1 \right)}-\frac{1}{\sqrt{n}}\sqrt{\left( \frac{1}{n}+1 \right)} \right]}{\left[ \sqrt{\left( \frac{1}{{{n}^{2}}}+1 \right)}-\frac{1}{n}\sqrt{\left( \frac{1}{n}+1 \right)} \right]}\]
\[=0\times \frac{\sqrt{0+1}-0.\sqrt{0+1}}{\sqrt{0+1}-0.\sqrt{0+1}}=0,\left[ \because n\to \infty \Rightarrow \frac{1}{\sqrt{n}}=0,\frac{1}{n}=0 \right]\]

 Question 09

Evaluate:
\[(i)\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{\sin x}}-1}{x}\]
\[(ii)\underset{x\to e}{\mathop{\lim }}\,\frac{{{\log }_{e}}x-1}{x-e}\]
\[(iii)\underset{x\to 0}{\mathop{\lim }}\,\frac{({{e}^{x}}-1){{\log }_{e}}(1+x)}{\sin x}\]

Solution:
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{\sin x}}-1}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{\sin x}}-1}{\sin x}\times \frac{\sin x}{x}\]
\[=\underset{\sin x\to 0}{\mathop{\lim }}\,\frac{{{e}^{\sin x}}-1}{\sin x}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x},[\because x\to 0\Rightarrow \sin x\to 0]\]
\[=1\times 1=1\]

Solution:
\[\underset{x\to e}{\mathop{\lim }}\,\frac{{{\log }_{e}}x-1}{x-e}=\underset{x\to e}{\mathop{\lim }}\,\frac{{{\log }_{e}}x-{{\log }_{e}}e}{x-e}\]
\[=\underset{x\to e}{\mathop{\lim }}\,\frac{{{\log }_{e}}\left( \frac{x}{e} \right)}{x-e},[let,x-e=z,then,x\to e\Rightarrow z\to 0,and,x=e+z]\]
\[=\underset{z\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}\left( 1+\frac{z}{e} \right)}{z}=\underset{z\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}\left( 1+\frac{z}{e} \right)}{\frac{z}{e}\times e}\]
\[=\frac{1}{e}\times \underset{\frac{z}{e}\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}\left( 1+\frac{z}{e} \right)}{\frac{z}{e}},\left[ \because z\to 0,\therefore \frac{z}{e}\to 0 \right]\]
\[=\frac{1}{e}\times 1=\frac{1}{e}\]

Solution:
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{({{e}^{x}}-1){{\log }_{e}}(1+x)}{\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{({{e}^{x}}-1)}{x}.\frac{{{\log }_{e}}(1+x)}{x}.{{x}^{2}}}{\frac{\sin x}{x}.x}\]
\[=\frac{\underset{x\to 0}{\mathop{\lim }}\,\frac{({{e}^{x}}-1)}{x}.\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}(1+x)}{x}.\underset{x\to 0}{\mathop{\lim }}\,x}{\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}}=\frac{1.1.0}{1}=0\]

Note: \[\underset{x\to a}{\mathop{\lim }}\,{{(f(x))}^{g(x)}}=\underset{x\to a}{\mathop{\lim }}\,{{e}^{g(x)\log f(x)}}\]
\[=\underset{x\to a}{\mathop{\lim }}\,{{e}^{g(x)\log [1+f(x)-1]}}=\underset{x\to a}{\mathop{\lim }}\,{{e}^{g(x)\log [f(x)-1]}}\]
\[\left[ \because \log (1+x)=x+\frac{{{x}^{2}}}{2}+….,|x|<1 \right]\]

 

 Question 10

Evaluate:
\[(i)\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{1+5{{x}^{2}}}{1+3{{x}^{2}}} \right)}^{\frac{1}{{{x}^{2}}}}}\]
\[(ii)\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{x-1+\cos x}{x} \right)}^{\frac{1}{x}}}\]

Solution:
\[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{1+5{{x}^{2}}}{1+3{{x}^{2}}} \right)}^{\frac{1}{{{x}^{2}}}}}={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1+5{{x}^{2}}}{1+3{{x}^{2}}}-1 \right).\frac{1}{{{x}^{2}}}}}\]
\[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{2{{x}^{2}}}{1+3{{x}^{2}}}.\frac{1}{{{x}^{2}}} \right)}}={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{2}{1+3{{x}^{2}}} \right)}}={{e}^{2}}\]

Solution:
\[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{x-1+\cos x}{x} \right)}^{\frac{1}{x}}}\]
\[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{x-1+\cos x}{x}-1 \right).\frac{1}{x}}}={{e}^{-\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos x}{{{x}^{2}}} \right)}}\]
\[={{e}^{-\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\frac{x}{2}}{{{x}^{2}}}}}={{e}^{-\frac{1}{2}{{\left( \underset{\frac{x}{2}\to 0}{\mathop{\lim }}\,\frac{\sin \frac{x}{2}}{\frac{x}{2}} \right)}^{2}}}}\]
\[={{e}^{-\frac{1}{2}\times {{1}^{2}}}}={{e}^{-\frac{1}{2}}}=\frac{1}{\sqrt{e}}\]

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