calculate-lim-x-0-sh-2sinx-sin-sh-2x-x-2-

Question Number 148568 by mathmax by abdo last updated on 29/Jul/21

calculate \lim_{x \to 0} \frac{sh (2 sinx) - \sin (sh (2 x))}{x^{2}}

Answered by Ar Brandon last updated on 29/Jul/21

L = \lim_{x \to 0} \frac{sh (2 \sin x) - \sin (sh (2 x))}{x^{2}}

= \lim_{x \to 0} \frac{2 \sin x - sh 2 x}{x^{2}} = \lim_{x \to 0} \frac{2 x - 2 x}{x^{2}} = 0

Answered by mathmax by abdo last updated on 29/Jul/21

u (x) = sh (2 sinx) - \sin (sh (2 x)) and v (x) = x^{2}

u^{^{'}} (x) = 2 cosxch (2 sinx) - 2 ch (2 x) \cos (sh (2 x))

\Rightarrow u^{(2)} (x) = - 2 sinxch (2 sinx) + {4 \cos}^{2} xsh (2 sinx) - 4 sh (2 x) \cos (sh (2 x))

+ {4 ch}^{2} (2 x) \sin (sh (2 x)) \Rightarrow \lim_{x \to 0} u^{(2)} (x) = 0

v (x) = x^{2} \Rightarrow v^{^{'}} (x) = 2 x and v^{(2)} (x) = 2 = \lim_{x \to 0} v^{(2)} (x) \Rightarrow

\lim_{x \to 0} \frac{u (x)}{v (x)} = \lim_{x \to 0} \frac{u^{(2)} (x)}{v^{(2)} (x)} = 0