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Question Number 62435 by mathsolverby Abdo last updated on 21/Jun/19

$$calculate{lim}_{x\to 0}\frac{{(1+x)}^{sinx}-1}{x^2}$$

Commented by Smail last updated on 22/Jun/19

$${(1+x)}^{sinx}=e^{ln\left({(1+x)}^{sinx}\right)}=e^{sinx\times ln(1+x)}$$$$ln(1+x)\sim xwhenxisnear0$$$$sinx\sim xwhenxisnear0$$$$so{(1+x)}^{sinx}\sim e^{x\times x}\sim e^{x^2}$$$$e^x\sim 1+xwhenxnear0$$$$so{(1+x)}^{sinx}\sim 1+x^2$$$$\underset{x\to 0}{lim}\frac{{(1+x)}^{sinx}-1}{x^2}=\underset{x\to 0}{lim}\frac{1+x^2-1}{x^2}$$$$=1$$

Commented by mathmax by abdo last updated on 23/Jun/19

$$letusehospitaltheoremu\left(x\right)={(1+x)}^{sinx}-1\Rightarrow u\left(x\right)=e^{sinxln(1+x)}-1\Rightarrow$$$$u^{{}^{\prime }}\left(x\right)=(cosxln(1+x)+\frac{sinx}{1+x})e^{sinxln(1+x)}\Rightarrow$$$$u^{\left(2\right)}\left(x\right)=\{-sinxln(1+x)+\frac{cosx}{1+x}+\frac{(1+x)cosx-sinx}{{(1+x)}^2})e^{sinxln(1+x)}$$$$+{(cosxln(1+x)+\frac{sinx}{1+x})}^2{e}^{sinxln(1+x)}\Rightarrow {lim}_{x\to 0}{u}^{\left(2\right)}\left(x\right)=2$$$$v\left(x\right)=x^2\Rightarrow g^{{}^{\prime }}\left(x\right)=2xand{v}^{\left(2\right)}\left(x\right)=2={lim}_{x\to 0}{g}^{\left(2\right)}\left(x\right)\Rightarrow$$$${lim}_{x\to 0}\frac{{(1+x)}^{sinx}-1}{x^2}={lim}_{x\to 0}\frac{u^{\left(2\right)}\left(x\right)}{v^{\left(2\right)}\left(x\right)}=\frac{2}{2}=1.$$

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