Search any question & find its solution
Question:
Answered & Verified by Expert
$\lim _{x \rightarrow 0} \frac{x^2 \cos x}{1-\cos x}$ is equal to
Options:
Solution:
2267 Upvotes
Verified Answer
The correct answer is:
2
2
Since, $\lim _{x \rightarrow 0} \frac{x^2 \cos x}{1-\cos x}=\lim _{x \rightarrow 0} \frac{x^2 \cos x}{1-\left(1-2 \sin ^2 \frac{x}{2}\right)}$
$$
=\lim _{x \rightarrow 0} \frac{x^2 \cos x}{2 \sin ^2 \frac{x}{2}}=2 \lim _{x \rightarrow 0} \cos x \times \lim _{x \rightarrow 0} \frac{\left(\frac{x}{2}\right)^2}{\sin ^2 \frac{x}{2}}=2.1=2
$$
$$
=\lim _{x \rightarrow 0} \frac{x^2 \cos x}{2 \sin ^2 \frac{x}{2}}=2 \lim _{x \rightarrow 0} \cos x \times \lim _{x \rightarrow 0} \frac{\left(\frac{x}{2}\right)^2}{\sin ^2 \frac{x}{2}}=2.1=2
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.