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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)))
$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{{ln}\left({x}+{e}^{{sinx}} \right)\:−{x}^{\mathrm{2}} }{{sh}\left(\mathrm{2}{x}\right)} \\ $$
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
$${let}\:{use}\:{hospital}\:{theorem}\:{let}\:{f}\left({x}\right)={ln}\left({x}+{e}^{{sinx}} \right)−{x}^{\mathrm{2}} \\ $$$${and}\:{g}\left({x}\right)={sh}\left(\mathrm{2}{x}\right)\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\frac{\mathrm{1}+{cosxe}^{{sinx}} }{{x}+{e}^{{sinx}} }\:−\mathrm{2}{x}\:\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\mathrm{2} \\ $$$${g}^{'} \left({x}\right)=\mathrm{2}{ch}\left(\mathrm{2}{x}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {g}^{'} \left({x}\right)=\mathrm{2}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{ln}\left({x}+{e}^{{sinx}} \right)−{x}^{\mathrm{2}} }{{sh}\left(\mathrm{2}{x}\right)}\:=\mathrm{1} \\ $$

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