lim_(x→0) ((sin x − x + (1/6) x^3 )/x^5 ) = ? ( Without L′Hospital , Taylor or Maclaurin Series ) .


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Question Number 157191 by naka3546 last updated on 20/Oct/21

$\lim_{x \to 0} \frac{\sin x - x + \frac{1}{6} x^{3}}{x^{5}} = ?$ $(W i t h o u t L^{'} H o s p i t a l, T a y l o r o r M a c l a u r i n S e r i e s) .$
	Answered by puissant last updated on 20/Oct/21

	$W e h a v e x - \frac{x^{3}}{3!} ⩽ s i n x ⩽ x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!}$ $\Rightarrow 0 ⩽ s i n x - x + \frac{x^{3}}{6} ⩽ \frac{x^{5}}{120}$ $\Rightarrow 0 ⩽ \frac{s i n x - x + \frac{1}{6} x^{3}}{x^{5}} ⩽ \frac{x^{5}}{120 x^{5}} = \frac{1}{120}$ $\Rightarrow \lim_{x \to 0} \frac{s i n x - x + \frac{x^{3}}{6}}{x^{5}} = \frac{1}{120} . .$
	Commented by naka3546 last updated on 20/Oct/21

	$T h a n k y o u .$
	Commented by Le_Professeur last updated on 20/Oct/21

	$w i t h o u t t a y l o r o r m a c l a u r i n s e r i e s (w i t h o u t c 7 s a n s u t i l i s e r)$
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