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Question Number 77856 by walidMTS last updated on 11/Jan/20

$$\underset{x\to 0}{lim}\left(\frac{e^x-x-1}{x}\right)$$

Commented by mathmax by abdo last updated on 11/Jan/20

$$wehave{e}^x=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\Rightarrow$$$$e^x=1+x+\frac{x^2}{2}+o\left(x^3\right)(x\to 0)\Rightarrow e^x-1-x=\frac{x^2}{2}+o\left(x^3\right)\Rightarrow$$$$\frac{e^x-x-1}{x}=\frac{x}{2}+o\left(x^2\right)\Rightarrow {lim}_{x\to 0}\frac{e^x-x-1}{x}={lim}_{x\to 0}\frac{x}{2}=0$$

Answered by Rio Michael last updated on 11/Jan/20

$$\underset{x\to 0}{\lim }\left(\frac{e^x-x-1}{x}\right)=\frac{\underset{x\to 0}{\lim }(e^x-x-1)}{\underset{x\to 0}{\lim }x}$$$$=\underset{x\to 0}{\lim }{e}^x-1=0$$

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