lim _{x rightarrow 0} frac{1-cos (1-cos x)}{s | vedclass

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

Question

(D) Given that,$\lim _{x ightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x}$ $= \lim _{x}$ ${ightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} 	i

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(D) Given that,$\lim _{x ightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x}$ $= \lim _{x}$ ${ightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} 	imes \left(\frac{1-\cos x}{x^2}ight)^2 	imes \left(\frac{x}{\sin x}ight)^4$ As $x ightarrow 0$,$(1-\cos x) ightarrow 0$. Using the standard limit $\lim _{	heta ightarrow 0} \frac{1-\cos 	heta}{	heta^2} = \frac{1}{2}$ and $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$: $= \frac{1}{2} 	imes \left(\frac{1}{2}ight)^2 	imes (1)^4$ $= \frac{1}{2} 	imes \frac{1}{4} 	imes 1 = \frac{1}{8}$

$\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x} = $

Similar Questions

Let $f(x) = \frac{1}{3} x \sin x - (1 - \cos x)$. The smallest positive integer $k$ such that $\lim_{x \rightarrow 0} \frac{f(x)}{x^k} \neq 0$ is

The value of $\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + bx + 4}}{{{x^2} + ax + 5}}} \right)$ is

$\lim _{x \rightarrow 1} (1 + \log _{e} x)^{1 / \log _{e} x}$ is equal to

$\lim _{x \rightarrow \infty}\left(\frac{x+8}{x+1}\right)^{x+5} = \dots$

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n k^2 x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module