JCECE 2025 - Counselling Registration (Started), Seat Allotment, Choice Filling, Merit List, Cutoff

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.

  • 642 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 )

If $a=\lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ and $b=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$, then the value of $a b^3$ is :

$\lim _{x \rightarrow 0} \frac{\cos 2 x-\cos 4 x}{x^2}$ equals

$\lim _{x \rightarrow 0} \frac{\sin 8 x}{\tan 3 x}$ equals

GNA University B.Tech Admissions 2025

100% Placement Assistance | Avail Merit Scholarships | Highest CTC 43 LPA

UPES B.Tech Admissions 2025

Ranked #42 among Engineering colleges in India by NIRF | Highest Package 1.3 CR , 100% Placements | Last Date to Apply: 15th July

$\begin{matrix} lim\\x \to 0\end{matrix}\frac{sin 3x+sin x}{x}=$

$\lim_{\theta \to \frac{\pi }{2}}\left ( \sec \theta - \tan \theta \right )$

JEE Main 2025 College Predictor
Know your college admission chances in NITs, IIITs and CFTIs, many States/ Institutes based on your JEE Main rank by using JEE Main 2025 College Predictor.
Use Now

$\lim_{x\to1 } \left ( 1+ \cos\pi x \right )\cot ^{2}\pi x$

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.

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

As $\lim _{x \rightarrow 0} \frac{\tan x}{x}=\lim _{x \rightarrow 0} \frac{\sin x}{x} \times \frac{1}{\cos x}$

$
=\lim _{x \rightarrow 0} \frac{\sin x}{x} \times \lim _{x \rightarrow 0} \frac{1}{\cos x}=1 \times 1
$

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

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

$
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{\sin h}{h} \\
& =1
\end{aligned}
$

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

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

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

As $\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x}=\lim _{y \rightarrow 0} \frac{y}{\sin y} \quad\left[\because \sin ^{-1} x=y\right]$
$=1$
(viii) $\lim _{x \rightarrow 0} \frac{\tan ^{-1} x}{x}=1$

Study it with Videos

Trigonometric Limits

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

Back to top