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

calculate lim_(x→0) ((sh(2sinx)−sin(sh(2x)))/x^2 )

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

L=lim_(x→0) ((sh(2sinx)−sin(sh(2x)))/x^2 ) =lim_(x→0) ((2sinx−sh2x)/x^2 )=lim_(x→0) ((2x−2x)/x^2 )=0

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

u(x)=sh(2sinx)−sin(sh(2x)) and v(x)=x^2 u^′ (x)=2cosxch(2sinx)−2ch(2x)cos(sh(2x)) ⇒u^((2)) (x)=−2sinxch(2sinx)+4cos^2 xsh(2sinx)−4sh(2x)cos(sh(2x)) +4ch^2 (2x) sin(sh(2x)) ⇒lim_(x→0) u^((2)) (x)=0 v(x)=x^2 ⇒v^′ (x)=2x and v^((2)) (x)=2 =lim_(x→0) v^((2)) (x) ⇒ lim_(x→0) ((u(x))/(v(x)))=lim_(x→0) ((u^((2)) (x))/(v^((2)) (x)))=0