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 evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{1 - \cos x}}{x}$.Using the trigonometric identity $1 - \cos x = 2 \sin^2(x/2)$,the

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) We evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{1 - \cos x}}{x}$.Using the trigonometric identity $1 - \cos x = 2 \sin^2(x/2)$,the expression becomes:$\mathop {\lim }\limits_{x 	o 0} \frac{2 \sin^2(x/2)}{x}$.Multiply and divide by $x/2$ to use the standard limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$:$\mathop {\lim }\limits_{x 	o 0} \left( \frac{2 \sin^2(x/2)}{x} \cdot \frac{x/4}{x/4} ight) = \mathop {\lim }\limits_{x 	o 0} \left( 2 \cdot \left( \frac{\sin(x/2)}{x/2} ight)^2 \cdot \frac{x}{4} ight) = 2 \cdot (1)^2 \cdot 0 = 0$.Alternatively,using $L$-Hospital's rule:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}(1 - \cos x)}{\frac{d}{dx}(x)} = \mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{1} = \sin(0) = 0$.

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

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \,\frac{{x\,\cot \,\left( {4x} \right)}}{{{{\sin }^2}\,x\,{{\cot }^2}\,\left( {2x} \right)}}$ is equal to

$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)}{x \tan 2 x+\frac{2 x}{3} \tan 3 x} = $

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

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

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\tan 3x}}{x} + \cos x} \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module