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find-lim-x-0-ln-x-e-sinx-x-2-sh-2x-




Question Number 37816 by prof Abdo imad last updated on 17/Jun/18
find lim_(x→0)     ((ln(x+e^(sinx) ) −x^2 )/(sh(2x)))
findlimx0ln(x+esinx)x2sh(2x)
Commented by math khazana by abdo last updated on 28/Jun/18
let use hospital theorem let f(x)=ln(x+e^(sinx) )−x^2   and g(x)=sh(2x) ⇒  f^′ (x)=((1+cosxe^(sinx) )/(x+e^(sinx) )) −2x  ⇒lim_(x→0) f(x)=2  g^′ (x)=2ch(2x) ⇒lim_(x→0) g^′ (x)=2 ⇒  lim_(x→0)      ((ln(x+e^(sinx) )−x^2 )/(sh(2x))) =1
letusehospitaltheoremletf(x)=ln(x+esinx)x2andg(x)=sh(2x)f(x)=1+cosxesinxx+esinx2xlimx0f(x)=2g(x)=2ch(2x)limx0g(x)=2limx0ln(x+esinx)x2sh(2x)=1

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