lim _{x rightarrow 0} frac{1-cos x}{x^2} = | vedclass

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

Question

(C) We know that the limit is of the form $\frac{0}{0}$ as $x \rightarrow 0$. Using the trigonometric identity $1 - \cos x = 2 \sin^2(\frac{x}{2})$,th

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) We know that the limit is of the form $\frac{0}{0}$ as $x ightarrow 0$. Using the trigonometric identity $1 - \cos x = 2 \sin^2(\frac{x}{2})$,the expression becomes: $\lim _{x ightarrow 0} \frac{2 \sin^2(\frac{x}{2})}{x^2}$ $= 2 \lim _{x ightarrow 0} \left( \frac{\sin(\frac{x}{2})}{x} ight)^2$ $= 2 \lim _{x ightarrow 0} \left( \frac{\sin(\frac{x}{2})}{2 \cdot \frac{x}{2}} ight)^2$ $= 2 \cdot \frac{1}{4} \lim _{x ightarrow 0} \left( \frac{\sin(\frac{x}{2})}{\frac{x}{2}} ight)^2$ Since $\lim _{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$,we get: $= 2 \cdot \frac{1}{4} \cdot (1)^2 = \frac{1}{2}$.

$\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2} = $

Similar Questions

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{\sin bx}$,where $a, b \neq 0$.

$\mathop {\lim }\limits_{h \to 0} \frac{{2\left[ {\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)} \right]}}{{\sqrt 3 h(\sqrt 3 \cos h - \sin h)}} = $

$\mathop {\lim }\limits_{x \to 2} \left( {\frac{{\sqrt {1 - \cos \{ 2(x - 2)\} } }}{{x - 2}}} \right) = $

If $\mathop {\lim }\limits_{x \to 0} kx\,\text{cosec}\,x = \mathop {\lim }\limits_{x \to 0} x\,\text{cosec}\,kx$,then $k = $

$\lim _{x \rightarrow 0} \frac{\cos (m x)-\cos (n x)}{x^2} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module