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Question Number 33313 by abdo imad last updated on 14/Apr/18

$$find{lim}_{x\to 0}\frac{ln(1+sinx)-sin\left(ln(1+x)\right)}{x^2}$$

Commented by abdo imad last updated on 17/Apr/18

$$letputu\left(x\right)=ln(1+sinx)-sin\left(ln(1+x)\right)andv\left(x\right)=x^2$$$$letusehospitaltheorem$$$${lim}_{x\to 0}\frac{u\left(x\right)}{v\left(x\right)}={lim}_{x\to 0}\frac{u^{{}^{″}}\left(x\right)}{v^{{}^{″}}\left(x\right)}butwehave$$$$u^{{}^{\prime }}\left(x\right)=\frac{cosx}{1+sinx}-\frac{1}{1+x}cos\left(ln(1+x)\right)\Rightarrow$$$$u^{{}^{″}}\left(x\right)=\frac{-sinx(1+sinx)-{cos}^{2}x}{{(1+sinx)}^2}-(\frac{-1}{{(1+x)}^2}cos\left(ln(1+x)\right)$$$$-\frac{1}{1+x}\frac{1}{1+x}sin\left(ln(1+x)\right)\Rightarrow {lim}_{x\to 0}{u}^{{}^{″}}\left(x\right)=-1+1=0$$$$alsowehave{v}^{{}^{\prime }}\left(x\right)=2x\Rightarrow v^{{}^{″}}\left(x\right)=2\Rightarrow$$$${lim}_{x\to 0}\frac{u\left(x\right)}{v\left(x\right)}=0$$

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