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

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Question

(A) We use the Taylor series expansion for $\sin x$: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = x - \frac{x^3}{6} + \frac{x^5}{120} - \do

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$1/120$

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(A) We use the Taylor series expansion for $\sin x$: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$ Substituting this into the limit expression: $\mathop {\lim }\limits_{x 	o 0} \frac{(x - \frac{x^3}{6} + \frac{x^5}{120} - \dots) - x + \frac{x^3}{6}}{x^5}$ $= \mathop {\lim }\limits_{x 	o 0} \frac{\frac{x^5}{120} - \dots}{x^5}$ $= \frac{1}{120}$.

$\mathop {\lim }\limits_{x \to 0} \left( \frac{\sin x - x + \frac{x^3}{6}}{x^5} \right) = $

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