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.

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.

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.

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

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.

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

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

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

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

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

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

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