mathop {lim }limits_{x to 0} frac{{tan x - si | vedclass

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Question

(A) We have $\mathop {\lim }\limits_{x 	o 0} \frac{{	an x - \sin x}}{{{x^3}}}$.Using the expansion $	an x = x + \frac{x^3}{3} + \dots$ and $\sin x

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) We have $\mathop {\lim }\limits_{x 	o 0} \frac{{	an x - \sin x}}{{{x^3}}}$.Using the expansion $	an x = x + \frac{x^3}{3} + \dots$ and $\sin x = x - \frac{x^3}{6} + \dots$,we get:$\mathop {\lim }\limits_{x 	o 0} \frac{{(x + \frac{x^3}{3} + \dots) - (x - \frac{x^3}{6} + \dots)}}{{{x^3}}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{x^3}{3} + \frac{x^3}{6}}}{{{x^3}}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{3x^3}{6}}}{{{x^3}}} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{x^3}}} = $

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