The value of the limit lim_{x rightarrow 0} l | vedclass

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Question

(A) Let $p = \lim_{x \rightarrow 0} \left(\frac{x}{\sin x}\right)^{6/x^2}$.Taking the natural logarithm on both sides:$\ln p = \lim_{x \rightarrow 0}

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$e$

Vedclass · Answer

(A) Let $p = \lim_{x \rightarrow 0} \left(\frac{x}{\sin x}\right)^{6/x^2}$.Taking the natural logarithm on both sides:$\ln p = \lim_{x \rightarrow 0} \frac{6}{x^2} \ln \left(\frac{x}{\sin x}\right)$.Using the Taylor series expansion for $\sin x = x - \frac{x^3}{6} + O(x^5)$:$\frac{x}{\sin x} = \frac{x}{x - \frac{x^3}{6} + O(x^5)} = \left(1 - \frac{x^2}{6} + O(x^4)\right)^{-1} = 1 + \frac{x^2}{6} + O(x^4)$.Now,$\ln p = \lim_{x \rightarrow 0} \frac{6}{x^2} \ln \left(1 + \frac{x^2}{6} + O(x^4)\right)$.Using the expansion $\ln(1+u) = u - \frac{u^2}{2} + \dots$ for $u = \frac{x^2}{6}$:$\ln p = \lim_{x \rightarrow 0} \frac{6}{x^2} \left(\frac{x^2}{6} + O(x^4)\right) = \lim_{x \rightarrow 0} (1 + O(x^2)) = 1$.Since $\ln p = 1$,we have $p = e^1 = e$.

The value of the limit $\lim_{x \rightarrow 0} \left(\frac{x}{\sin x}\right)^{6/x^2}$ is

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