Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 24 January, Morning Shift — Question 7

$\lim _{x \rightarrow 0} \operatorname{cosec} x\left(\sqrt{2 \cos ^{2} x+3 \cos x}-\sqrt{\cos ^{2} x+\sin x+4}\right)$ is

Option A:
0
Option B:
$\frac{1}{2 \sqrt{5}}$
Option C:
$\frac{1}{\sqrt{15}}$
Option D:
$-\frac{1}{2 \sqrt{5}}$
Correct

Answer: D

Step-by-step solution

$\lim _{x \rightarrow 0} \operatorname{cosec}\left(\sqrt{2 \cos ^{2} x+3 \cos x}-\sqrt{\cos ^{2} x+\sin x+4}\right)$

\lim _{x \rightarrow 0} \frac{\operatorname{cosec} x\left(\cos ^{2} x+3 \cos x-\sin x-4\right)}{\left(\sqrt{2 \cos ^{2} x+3 \cos x}+\sqrt{\cos ^{2} x+\sin x+4}\right)}

\lim _{x \rightarrow 0} \frac{1}{\sin x} \frac{\left(\cos ^{2} x+3 \cos x-4\right)-\sin x}{\left(\sqrt{2 \cos ^{2} x+3 \cos x}+\sqrt{\cos ^{2} x+\sin x+4}\right)}

\lim _{x \rightarrow 0} \frac{(\cos x+4)(\cos x-1)-\sin x}{\sin x\left(\sqrt{2 \cos ^{2} x+3 \cos x}+\sqrt{\cos ^{2} x+\sin x+4}\right)}

\lim _{x \rightarrow 0} \frac{-2 \sin ^{2} \frac{x}{2}(\cos x+4)-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}\left(\sqrt{2 \cos ^{2} x+3 \cos x}+\sqrt{\cos ^{2} x+\sin x+4}\right)}

\lim _{x \rightarrow 0} \frac{-\left(\sin \frac{x}{2}(\cos x+4)+\cos \frac{x}{2}\right)}{\cos \frac{x}{2}\left(\sqrt{2 \cos ^{2} x+3 \cos x}+\sqrt{\cos ^{2} x+\sin x+4}\right)}

-\frac{1}{2 \sqrt{5}}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Evaluation of Limit of Functions

Let f: mathbbR-\0\ arrow mathbbR be a function such that f(x)-6 f (1/x )=35/3 x-5/2 . If the lim x arrow 0 (1/alpha x+f(x) )=beta ; alpha…