mathop {lim }limits_{x to 0} frac{{1 - cos x} | vedclass

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

Question

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{1 - \cos x}}{{{{\sin }^2}x}}$.Using the trigonometric identity $1 - \cos x

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{1 - \cos x}}{{{{\sin }^2}x}}$.Using the trigonometric identity $1 - \cos x = 2 \sin^2(\frac{x}{2})$ and $\sin^2 x = (2 \sin(\frac{x}{2}) \cos(\frac{x}{2}))^2 = 4 \sin^2(\frac{x}{2}) \cos^2(\frac{x}{2})$:$\mathop {\lim }\limits_{x 	o 0} \frac{{2 \sin^2(\frac{x}{2})}}{{4 \sin^2(\frac{x}{2}) \cos^2(\frac{x}{2})}} = \mathop {\lim }\limits_{x 	o 0} \frac{1}{2 \cos^2(\frac{x}{2})}$.As $x 	o 0$,$\cos(\frac{x}{2}) 	o \cos(0) = 1$.Therefore,the limit is $\frac{1}{2(1)^2} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{{{\sin }^2}x}} = $

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{5\sin x + x\cos x}}{{2\tan x - {x^2}}}$ is

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

If $a > 0$ and $b < 0$,then $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt {1 - \cos 2ax} }}{{\sin bx}}$ is equal to:

$\lim _{x \rightarrow 0} \frac{\tan ^4 x-\sin ^4 x}{x^6} = $

$\lim _{t}$ ${\rightarrow 0}\left(1^{\frac{1}{\sin ^2 t}}+2^{\frac{1}{\sin ^2 t}}+\ldots +n^{\frac{1}{\sin ^2 t}}\right)^{\sin ^2 t}$ is equal to $.......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module