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

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Question

(D) Using the Taylor series expansion for $\sin x$: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$ Substituting this into the limit: $\mathop

Vedclass · Accepted Answer

$-\frac{1}{6}$

Vedclass · Answer

(D) Using the Taylor series expansion for $\sin x$: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$ Substituting this into the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots) - x}{x^3}$ $= \mathop {\lim }\limits_{x 	o 0} \frac{-\frac{x^3}{6} + \frac{x^5}{120} - \dots}{x^3}$ $= \mathop {\lim }\limits_{x 	o 0} (-\frac{1}{6} + \frac{x^2}{120} - \dots)$ $= -\frac{1}{6}$ Alternatively,applying $L$-Hospital's rule three times gives the same result.

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

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