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Limits of Trigonometric Functions - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Trigonometric Limits is considered one the most difficult concept.

  • 613 Questions around this concept.

Solve by difficulty

\lim_{x\rightarrow \frac{\pi }{2}}\: \: \frac{\cot x-\cos x}{\left ( \pi -2x \right )^{3}}equals:

\lim_{x\rightarrow 0}\frac{\sin \left ( \pi \cos ^{2}x \right )}{x^{2}}   is equal to :

if f(x) is continuous and f(\frac{9}{2})=\frac{2}{9}, then \lim_{x\rightarrow 0}f\left ( \frac{1-\cos 3x}{x^{2}} \right )  is equal to :

\lim_{x\rightarrow 2}\left ( \frac{\sqrt{1-\cos \left \{ 2(x-2) \right \}}}{x-2} \right )

Concepts Covered - 1

Trigonometric Limits

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.

\mathbf{\text{(i)}\;\;\;\lim_{x\rightarrow 0}\frac{\sin x}{x}=1}

\\\mathbf{\text{(ii)}\;\;\;\lim_{x\rightarrow 0}\frac{\tan x}{x}=1} \\\\\text{\;\;\;\;\;\;\;\;As}\;\;\lim_{x\rightarrow 0}\frac{\tan x}{x}=\lim_{x\rightarrow 0}\frac{\sin x}{x}\times\frac{1}{\cos x}\\\\\text{\;\;\;\;\;\;\;\;}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\lim_{x\rightarrow 0}\frac{\sin x}{x}\times\lim_{x\rightarrow 0}\frac{1}{\cos x}=1\times 1
\mathbf{\text{(iii)}\;\;\;\lim_{x\rightarrow a}\frac{\sin (x-a)}{x-a}=1}\\\\\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;As}\;\;\lim_{x\rightarrow a}\frac{\sin (x-a)}{x-a}=\lim_{h\rightarrow 0}\frac{\sin ((a+h)-a)}{(a+h)-a}\\\\\text{\;\;\;\;\;\;\;\;}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\lim_{h\rightarrow 0}\frac{\sin h}{h}\\\\\text{\;\;\;\;\;\;\;\;}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=1

\mathbf{\text{(iv)}\;\;\;\lim_{x\rightarrow a}\frac{\tan (x-a)}{x-a}=1}

\\(v)\;\;\lim_{x \rightarrow a} \frac{\sin (f(x))}{f(x)}=1,\;\;if\;\lim_{x\rightarrow a}f(x)=0\\ \,\,\,\,\,\,\,\:\:Similarly, \;\;\;\lim_{x\rightarrow a}\frac{\tan(f(x))}{f(x)}=1,\;\;if\;\lim_{x\rightarrow a}f(x)=0

\mathbf{\text{(vi)}\;\;\;\lim_{x\rightarrow 0}\cos x=1}

\\\mathbf{\text{(vii)}\;\;\;\lim_{x\rightarrow 0}\frac{\sin^{-1} x}{x}=1}\\ \\\text{\;\;\;\;\;\;\;\;\;As}\;\;\lim_{x\rightarrow 0}\frac{\sin^{-1} x}{x}=\lim_{y\rightarrow 0}\frac{y}{\sin y}\;\;\;\;\;\;\;\;[\because \sin^{-1}x=y]\\\\\text{\;\;\;\;\;\;\;\;}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=1

\\\mathbf{\text{(viii)}\;\;\;\lim_{x\rightarrow 0}\frac{\tan^{-1} x}{x}=1}

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Trigonometric Limits

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