# Limits, Continuity and Differentiability – Standard Results on Limits

## Limits, Continuity and Differentiability – Standard Results on Limits

To evaluate trigonometric limits, reduce the terms of the function in terms of sinθ and cosθ. Remove the positive and negative signs in between two terms i.e., express the function in product form. Arrange terms and take help of the following standard results.

Trigonometric Limits:

i) $$\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sin x}$$,

ii) $$\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}=1=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\tan x}$$,

iii) $$\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{-1}}x}{x}=1=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{{{\sin }^{-1}}x}$$,

iv) $$\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\tan }^{-1}}x}{x}=1=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{{{\tan }^{-1}}x}$$,

v) $$\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin {{x}^{\circ }}}{x}=\frac{\pi }{180}$$,

vi) $$\underset{x\to 0}{\mathop{\lim }}\,\cos x=1$$,

vii) $$\underset{x\to a}{\mathop{\lim }}\,\frac{\sin \left( x-a \right)}{x-a}=1$$,

viii) $$\underset{x\to a}{\mathop{\lim }}\,\frac{\tan \left( x-a \right)}{x-a}=1$$,

ix) $$\underset{x\to a}{\mathop{\lim }}\,{{\sin }^{-1}}x={{\sin }^{-1}}a,\,|a|\,\le \,1$$,

x) $$\underset{x\to a}{\mathop{\lim }}\,{{\cos }^{-1}}x={{\cos }^{-1}}a,\,|a|\,\le \,1$$,

xi) $$\underset{x\to a}{\mathop{\lim }}\,{{\tan }^{-1}}x={{\tan }^{-1}}a,\,-\infty <a<\infty$$,

xii) $$\underset{x\to \infty }{\mathop{\lim }}\,\frac{\sin x}{x}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{\cos x}{x}=0$$,

xiii) $$\underset{x\to \infty }{\mathop{\lim }}\,\frac{\sin 1/x}{1/x}=1$$,

Example: $$\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{\tan x}}-{{e}^{x}}}{\tan x-x}$$ is equal to

Solution: Since, $${{e}^{\tan x}}=1+\tan x+\frac{{{\tan }^{2}}x}{2!}+…$$,

But we know that, $$\tan x=x+\frac{{{x}^{3}}}{3}+\frac{2}{15}{{x}^{5}}+…$$,

∴ $${{e}^{\tan x}}=\left[ 1+\left\{ x+\frac{{{x}^{3}}}{3}+… \right\}+\frac{1}{2}{{\left\{ x+\frac{{{x}^{3}}}{3}+… \right\}}^{2}}+\frac{1}{6}{{\left\{ x+\frac{{{x}^{3}}}{3}+… \right\}}^{3}}+… \right]$$,

$$=1+\left( x+\frac{{{x}^{3}}}{3}+… \right)+\frac{1}{2}\left( {{x}^{2}}\frac{2}{3}{{x}^{4}}+… \right)+\frac{1}{6}\left( {{x}^{3}}+… \right)$$,

$$=1+x+\frac{{{x}^{2}}}{2}+{{x}^{3}}\left( \frac{1}{3}+\frac{1}{6} \right)+…$$,

$$=1+x+\frac{{{x}^{2}}}{2}+\frac{1}{2}{{x}^{3}}+…$$,

And $${{e}^{x}}=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+…$$,

∴ $${{e}^{\tan x}}-{{e}^{x}}={{x}^{3}}\left( \frac{1}{2}-\frac{1}{3} \right)+…$$,

$$=\frac{{{x}^{3}}}{6}+$$ Higher powers of x.

Similarly, $$\tan x-x=\frac{{{x}^{3}}}{3}+\frac{2}{15}{{x}^{5}}+$$ Higher powers of x

∴ $$\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{\tan x}}-{{e}^{x}}}{\tan x-x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{{{x}^{3}}}{6}+Higher\,\,powers\,\,of\,\,x}{\frac{{{x}^{3}}}{3}+Higher\,\,powers\,\,of\,\,x}=\frac{1}{2}$$.