The value of lim _{x rightarrow 0} left( frac | vedclass

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Question

(B) Let $L = \lim _{x \rightarrow 0} \frac{x}{(1-\sin x)^{1/8}-(1+\sin x)^{1/8}}$.Using the binomial expansion $(1+u)^n \approx 1+nu$ for small $u$,wh

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(B) Let $L = \lim _{x ightarrow 0} \frac{x}{(1-\sin x)^{1/8}-(1+\sin x)^{1/8}}$.Using the binomial expansion $(1+u)^n \approx 1+nu$ for small $u$,where $n = 1/8$:$(1-\sin x)^{1/8} \approx 1 - \frac{1}{8} \sin x$.$(1+\sin x)^{1/8} \approx 1 + \frac{1}{8} \sin x$.Substituting these into the limit expression:$L = \lim _{x ightarrow 0} \frac{x}{(1 - \frac{1}{8} \sin x) - (1 + \frac{1}{8} \sin x)}$.$L = \lim _{x ightarrow 0} \frac{x}{-\frac{2}{8} \sin x} = \lim _{x ightarrow 0} \frac{x}{-\frac{1}{4} \sin x}$.Since $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$,we have $\lim _{x ightarrow 0} \frac{x}{\sin x} = 1$.Therefore,$L = -4 	imes 1 = -4$.

The value of $\lim _{x \rightarrow 0} \left( \frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}} \right)$ is equal to:

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