mathop {lim }limits_{x to 0} frac{{sin left( | vedclass

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

Question

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\sin \left( {\pi {{\cos }^2}x} ight)}}{{{x^2}}}$Using the identity $\cos

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\sin \left( {\pi {{\cos }^2}x} ight)}}{{{x^2}}}$Using the identity $\cos^2 x = 1 - \sin^2 x$,we rewrite the expression:$= \mathop {\lim }\limits_{x 	o 0} \frac{{\sin \left( {\pi (1 - \sin^2 x)} ight)}}{{{x^2}}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{\sin \left( {\pi - \pi \sin^2 x} ight)}}{{{x^2}}}$Since $\sin(\pi - 	heta) = \sin 	heta$,we have:$= \mathop {\lim }\limits_{x 	o 0} \frac{{\sin \left( {\pi \sin^2 x} ight)}}{{{x^2}}}$Multiply and divide by $\pi \sin^2 x$:$= \mathop {\lim }\limits_{x 	o 0} \left( \frac{{\sin \left( {\pi \sin^2 x} ight)}}{{\pi \sin^2 x}} 	imes \frac{{\pi \sin^2 x}}{{{x^2}}} ight)$Using the standard limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{{\sin 	heta}}{	heta} = 1$ and $\mathop {\lim }\limits_{x 	o 0} \frac{{\sin x}}{x} = 1$:$= 1 	imes \pi 	imes (\mathop {\lim }\limits_{x 	o 0} \frac{{\sin x}}{x})^2 = 1 	imes \pi 	imes 1^2 = \pi$

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}}$ equals

Similar Questions

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

$\lim _{x \rightarrow 0} \frac{\cos 7 x^{\circ}-\cos 2 x^{\circ}}{x^2}$ is

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

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}$

$\lim _{x \rightarrow 0} \frac{\sin ^{2}\left(\pi \cos ^{4} x\right)}{x^{4}}$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module