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

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Question

(C) We evaluate the limit: $\lim _{x ightarrow 0} \frac{	an x - \sin x}{x^3}$.Since this is a $\frac{0}{0}$ form,we use the Taylor series expansion

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) We evaluate the limit: $\lim _{x ightarrow 0} \frac{	an x - \sin x}{x^3}$.Since this is a $\frac{0}{0}$ form,we use the Taylor series expansion:$	an x = x + \frac{x^3}{3} + O(x^5)$$\sin x = x - \frac{x^3}{6} + O(x^5)$Substituting these into the expression:$\lim _{x ightarrow 0} \frac{(x + \frac{x^3}{3}) - (x - \frac{x^3}{6})}{x^3} = \lim _{x ightarrow 0} \frac{\frac{x^3}{3} + \frac{x^3}{6}}{x^3} = \lim _{x ightarrow 0} \frac{\frac{x^3}{2}}{x^3} = \frac{1}{2}$.

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

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