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) Given the limit $ \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}} $. Substituting $ x = 0 $,we get the indeterminate form $ \left(\frac{0}{0}\right

Vedclass · Accepted Answer

$ \frac{1}{2} $

Vedclass · Answer

(C) Given the limit $ \lim _{x ightarrow 0} \frac{1-\cos x}{x^{2}} $. Substituting $ x = 0 $,we get the indeterminate form $ \left(\frac{0}{0}ight) $. Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $ x $: $ \lim _{x ightarrow 0} \frac{\frac{d}{dx}(1-\cos x)}{\frac{d}{dx}(x^{2})} = \lim _{x ightarrow 0} \frac{\sin x}{2x} $. Using the standard limit $ \lim _{x ightarrow 0} \frac{\sin x}{x} = 1 $,we get: $ \frac{1}{2} 	imes \lim _{x ightarrow 0} \frac{\sin x}{x} = \frac{1}{2} 	imes 1 = \frac{1}{2} $. Thus,the value of the limit is $ \frac{1}{2} $.

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

Similar Questions

The value of the limit of $\frac{x^3 - x^2 - 18}{x - 3}$ as $x$ tends to $3$ is

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$ is

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = $

$\lim _{x}$ ${\rightarrow 0} \frac{x+2 \sin x+3 \tan x-\tan ^3 x}{\sqrt{x^2+2 \sin x+\tan x+3}-\sqrt{\sin ^2 x-2 \tan x-x+3}} =$

The value of $\lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module