lim _{x rightarrow 0} left( frac{sin (pi cos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the limit: $\lim _{x \rightarrow 0} \frac{\sin (\pi \cos ^2 x)}{x^2}$ Since $\cos ^2 x = 1 - \sin ^2 x$,we have $\pi \cos ^2 x = \pi - \pi \

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(B) Given the limit: $\lim _{x ightarrow 0} \frac{\sin (\pi \cos ^2 x)}{x^2}$ Since $\cos ^2 x = 1 - \sin ^2 x$,we have $\pi \cos ^2 x = \pi - \pi \sin ^2 x$. Using the identity $\sin (\pi - 	heta) = \sin 	heta$,we get $\sin (\pi - \pi \sin ^2 x) = \sin (\pi \sin ^2 x)$. Now,the limit becomes: $\lim _{x ightarrow 0} \frac{\sin (\pi \sin ^2 x)}{x^2}$. Multiplying and dividing by $\pi \sin ^2 x$: $\lim _{x}$ ${ightarrow 0} \left( \frac{\sin (\pi \sin ^2 x)}{\pi \sin ^2 x} ight) 	imes \left( \frac{\pi \sin ^2 x}{x^2} ight)$. As $x ightarrow 0$,$\sin ^2 x ightarrow 0$,so $\frac{\sin (\pi \sin ^2 x)}{\pi \sin ^2 x} ightarrow 1$. Also,$\lim _{x ightarrow 0} \frac{\sin ^2 x}{x^2} = 1$. Therefore,the limit is $1 	imes \pi 	imes 1 = \pi$.

$\lim _{x \rightarrow 0} \left( \frac{\sin (\pi \cos ^2 x)}{x^2} \right) = $

Similar Questions

$\lim _{x \rightarrow 0} \left( \frac{\sin ax}{\tan bx} \right)$ is equal to

$\lim _{x \rightarrow 0} \frac{\tan ^3 x - \sin ^3 x}{x^5}$ is equal to

$\mathop {\lim }\limits_{\theta \to 0} \frac{{4\theta (\tan \theta - \sin \theta )}}{{{{(1 - \cos 2\theta )}^2}}}$ is

$\mathop {\lim }\limits_{x \to 0} \frac{x}{{\tan x}}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{3\sin x - \sin 3x}}{{{x^3}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module