lim _{x rightarrow 0} frac{sqrt{cos x} - sqrt | vedclass

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Question

(B) Let $L = \lim _{x \rightarrow 0} \frac{(\cos x)^{1/2} - (\cos x)^{1/3}}{\sin ^2 x}$.Using the Taylor series expansion for $\cos x \approx 1 - \fra

Vedclass · Accepted Answer

$-1/12$

Vedclass · Answer

(B) Let $L = \lim _{x \rightarrow 0} \frac{(\cos x)^{1/2} - (\cos x)^{1/3}}{\sin ^2 x}$.Using the Taylor series expansion for $\cos x \approx 1 - \frac{x^2}{2}$ and $\sin x \approx x$ as $x \rightarrow 0$:$L = \lim _{x \rightarrow 0} \frac{(1 - \frac{x^2}{2})^{1/2} - (1 - \frac{x^2}{2})^{1/3}}{x^2}$.Using the binomial expansion $(1+u)^n \approx 1 + nu$ for small $u$:$(1 - \frac{x^2}{2})^{1/2} \approx 1 - \frac{1}{2}(\frac{x^2}{2}) = 1 - \frac{x^2}{4}$.$(1 - \frac{x^2}{2})^{1/3} \approx 1 - \frac{1}{3}(\frac{x^2}{2}) = 1 - \frac{x^2}{6}$.Substituting these into the limit:$L = \lim _{x \rightarrow 0} \frac{(1 - \frac{x^2}{4}) - (1 - \frac{x^2}{6})}{x^2} = \lim _{x \rightarrow 0} \frac{-\frac{x^2}{4} + \frac{x^2}{6}}{x^2} = -\frac{1}{4} + \frac{1}{6} = \frac{-3 + 2}{12} = -\frac{1}{12}$.

$\lim _{x \rightarrow 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin ^2 x} = $

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