lim _{x rightarrow 0} frac{tan x - sin x}{x^3 | vedclass

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(C) We evaluate the limit $L = \lim _{x ightarrow 0} \frac{	an x - \sin x}{x^3}$.Since this is a $\frac{0}{0}$ indeterminate form,we apply $L$'Hosp

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$\frac{1}{2}$

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(C) We evaluate the limit $L = \lim _{x ightarrow 0} \frac{	an x - \sin x}{x^3}$.Since this is a $\frac{0}{0}$ indeterminate form,we apply $L$'Hospital's Rule:$L = \lim _{x ightarrow 0} \frac{\sec^2 x - \cos x}{3x^2}$.This is still a $\frac{0}{0}$ form. Applying $L$'Hospital's Rule again:$L = \lim _{x ightarrow 0} \frac{2 \sec^2 x 	an x + \sin x}{6x}$.Using the standard limits $\lim _{x ightarrow 0} \frac{	an x}{x} = 1$ and $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$:$L = \lim _{x}$ ${ightarrow 0} \left( \frac{2 \sec^2 x \cdot 	an x}{6x} + \frac{\sin x}{6x} ight) = \frac{2(1)(1)}{6} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

$\lim _{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$ is equal to

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