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

$\lim_{x \rightarrow -2^{+}} ([x]^2 - [x] - 2) + \lim_{x \rightarrow -3^{-}} ([x]^2 - 4[x] + 3) =$

The value of $\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4{x^2} + 5x + 8} }}{{4x + 5}}$ is

$\mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }}$ is equal to

Evaluate the limit: $\mathop {\lim }\limits_{x \to \infty } [x({a^{1/x}} - 1)]$,where $a > 1$.

If $f(x) = \frac{\sin(e^{x-2} - 1)}{\log(x-1)},$ then $\lim_{x \to 2} f(x)$ is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module