Search any question & find its solution
Question:
Answered & Verified by Expert
$\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$ equals
Options:
Solution:
2332 Upvotes
Verified Answer
The correct answer is:
$\pi$
$\pi$
Consider
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\sin \left(\pi-\pi \sin ^2 x\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{\pi \sin ^2 x} \times \frac{\left(\pi \sin ^2 x\right)}{x^2}=\pi
\end{aligned}
$$
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\sin \left(\pi-\pi \sin ^2 x\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{\pi \sin ^2 x} \times \frac{\left(\pi \sin ^2 x\right)}{x^2}=\pi
\end{aligned}
$$
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.