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Question Number 86603 by Ar Brandon last updated on 29/Mar/20

$$\underset{x\to -1}{\lim }\frac{e^x}{{(1+x)}^n}$$

Commented by abdomathmax last updated on 29/Mar/20

$$letf\left(x\right)=\frac{e^x}{{(1+x)}^n}changement1+x=ugive$$$$f\left(x\right)=\frac{e^{u-1}}{u^n}soifn>0{lim}_{x\to -1}f\left(x\right)={lim}_{u\to 0}\frac{e^{u-1}}{u^n}$$$$=\frac{e^{-1}}{0^{+}}=+\mathrm{\infty }$$$$ifn<0{lim}_{x\to -1}f\left(x\right)={lim}_{u\to 0}{u}^{-n}{e}^{u-1}=0$$

Commented by Ar Brandon last updated on 29/Mar/20

$$great!$$

Answered by Rio Michael last updated on 29/Mar/20

$$+\mathrm{\infty }?$$

Commented by Ar Brandon last updated on 29/Mar/20

$$how,please?$$

Answered by Rio Michael last updated on 29/Mar/20

$$\mathrm{my}\mathrm{try}$$$$\underset{x\to -1}{\lim }\frac{e^x}{{(1+x)}^n}=\underset{x\to -1}{\lim }\frac{e^x}{n{(1+x)}^{n-1}}=+\mathrm{\infty }\mathrm{for}n>0$$$$\mathrm{and}\underset{x\to -1}{\lim }\frac{e^x}{n{(1+x)}^{n-1}}=0\mathrm{for}n<0$$

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