Limits of Trigonometric Functions - Practice Questions & MCQ

Trigonometric Limits

In the trigonometric limit, apart from using the method of direct substitution, factorization and rationalization (same as given in algebraic limits), we can use the following formulae.

(i) $\underset{\mathbf{x}\to \mathbf{0}}{lim}\frac{\sin \mathrm{x}}{\mathbf{x}}=\mathbf{1}$
(ii) $\underset{\mathbf{x}\to \mathbf{0}}{lim}\frac{\tan \mathrm{x}}{\mathbf{x}}=\mathbf{1}$

As $\underset{x\to 0}{lim}\frac{\tan x}{x}=\underset{x\to 0}{lim}\frac{\sin x}{x}\times \frac{1}{\cos x}$

$=\underset{x\to 0}{lim}\frac{\sin x}{x}\times \underset{x\to 0}{lim}\frac{1}{\cos x}=1\times 1$

(iii) $\underset{\mathbf{x}\to \mathbf{a}}{lim}\frac{\sin (\mathbf{x}-\mathbf{a})}{\mathbf{x}-\mathbf{a}}=\mathbf{1}$

As $\underset{x\to a}{lim}\frac{\sin (x-a)}{x-a}=\underset{h\to 0}{lim}\frac{\sin ((a+h)-a)}{(a+h)-a}$

$\begin{array}{rl}& =\underset{h\to 0}{lim}\frac{\sin h}{h}\\ & =1\end{array}$

(iv) $\underset{\mathbf{x}\to \mathrm{a}}{lim}\frac{\tan (\mathbf{x}-\mathbf{a})}{\mathbf{x}-\mathbf{a}}=\mathbf{1}$
(v) $\underset{x\to a}{lim}\frac{\sin (f(x))}{f(x)}=1$, if $\underset{x\to a}{lim}f(x)=0$

Similarly, $\underset{x\to a}{lim}\frac{\tan (f(x))}{f(x)}=1$, if $\underset{x\to a}{lim}f(x)=0$
(vi) $\underset{x\to 0}{lim}\cos x=1$

(vii) $\underset{\mathrm{x}\to 0}{lim}\frac{{\sin }^{-1}\mathrm{x}}{\mathrm{x}}=1$

As $\underset{x\to 0}{lim}\frac{{\sin }^{-1}x}{x}=\underset{y\to 0}{lim}\frac{y}{\sin y}[\because {\sin }^{-1}x=y]$
$=1$
(viii) $\underset{x\to 0}{lim}\frac{{\tan }^{-1}x}{x}=1$