lim _{x} {rightarrow 0} frac{8}{x^8}left[1-co | vedclass

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

Question

(D) Let $L = \lim _{x}$ ${\rightarrow 0} \frac{8}{x^8}\left[1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}

Vedclass · Accepted Answer

$\frac{1}{32}$

Vedclass · Answer

(D) Let $L = \lim _{x}$ ${ightarrow 0} \frac{8}{x^8}\left[1-\cos \left(\frac{x^2}{2}ight)-\cos \left(\frac{x^2}{4}ight)+\cos \left(\frac{x^2}{2}ight) \cdot \cos \left(\frac{x^2}{4}ight)ight]$Factor the expression inside the bracket:$L = \lim _{x}$ ${ightarrow 0} \frac{8}{x^8}\left[\left(1-\cos \frac{x^2}{2}ight) - \cos \left(\frac{x^2}{4}ight)\left(1-\cos \frac{x^2}{2}ight)ight]$$L = \lim _{x ightarrow 0} \frac{8}{x^8}\left(1-\cos \frac{x^2}{2}ight)\left(1-\cos \frac{x^2}{4}ight)$Using the identity $1-\cos 	heta = 2 \sin^2 \left(\frac{	heta}{2}ight)$:$L = \lim _{x ightarrow 0} \frac{8}{x^8} \left(2 \sin^2 \frac{x^2}{4}ight) \left(2 \sin^2 \frac{x^2}{8}ight)$$L = \lim _{x ightarrow 0} \frac{32}{x^8} \left(\sin^2 \frac{x^2}{4}ight) \left(\sin^2 \frac{x^2}{8}ight)$Multiply and divide by $\left(\frac{x^2}{4}ight)^2$ and $\left(\frac{x^2}{8}ight)^2$:$L = 32 \lim _{x}$ ${ightarrow 0} \left(\frac{\sin \frac{x^2}{4}}{\frac{x^2}{4}}ight)^2 \left(\frac{x^4}{16}ight) \left(\frac{\sin \frac{x^2}{8}}{\frac{x^2}{8}}ight)^2 \left(\frac{x^4}{64}ight) \cdot \frac{1}{x^8}$$L = 32 \cdot (1)^2 \cdot \frac{1}{16} \cdot (1)^2 \cdot \frac{1}{64}$$L = 32 \cdot \frac{1}{1024} = \frac{1}{32}$

$\lim _{x}$ ${\rightarrow 0} \frac{8}{x^8}\left[1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^2}{4}\right)\right]$ is equal to

Similar Questions

$\mathop {\lim }\limits_{x \to 0} x^2(1+2+3+...+[\frac{1}{|x|}])$ is equal to (where $[.]$ denotes the greatest integer function).

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^5} - 1}}{{{{(1 + x)}^3} - 1}} = $

$\lim _{n \rightarrow \infty}\left(1+\frac{1+\frac{1}{2}+\ldots+\frac{1}{n}}{n^{2}}\right)^{n} = \dots$

The value of $\mathop {\lim }\limits_{n \to \infty } \frac{{{x^n}}}{{{x^n} + 1}}$ where $x < -1$ is

If $[x]$ is the greatest integer function,then $\lim _{x \rightarrow 3^{-}} \frac{(3-|x|+\sin |3-x|) \cos [9-3 x]}{|3-x|[3 x-9]} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module