lim_(x→0) ((sinh (2x)−sin 2x)/x^5 ) =?


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Question Number 114696 by bemath last updated on 20/Sep/20

$\lim_{x \to 0} \frac{\sinh (2 x) - \sin 2 x}{x^{5}} = ?$
	Answered by bobhans last updated on 20/Sep/20

	$e^{2 x} = 1 + 2 x + 2 x^{2} + \frac{4 x^{3}}{3} + \frac{2 x^{4}}{3} + \frac{32 x^{5}}{120} + . . .$ $e^{2 x} = 1 - 2 x + 2 x^{2} - \frac{4 x^{3}}{3} + \frac{2 x^{4}}{3} - \frac{32 x^{5}}{120} + . . .$ $\sinh 2 x = \frac{1}{2} (4 x + \frac{8 x^{3}}{3} + \frac{64 x^{5}}{120} + . . .)$ $= 2 x + \frac{4 x^{3}}{3} + \frac{32 x^{5}}{120} + . . .$ $\lim_{x \to 0} \frac{\sinh 2 x - \sin 2 x}{x^{5}} = \lim_{x \to 0} \frac{(2 x + \frac{4 x^{3}}{3} + \frac{32 x^{5}}{120}) - (2 x - \frac{4 x^{3}}{3} + \frac{32 x^{5}}{120})}{x^{5}} =$ $\lim_{x \to 0} \frac{\frac{8 x^{3}}{3}}{x^{5}} = 0$
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