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I need help understanding this proof:

Prove that $$\lim_{x\to 0}\frac{e^x-1}{x}=1.$$

For $x>0$ and $n\in\Bbb N$: $$1\leq\frac{(1+\frac {x}{n})^n -1}{x}=\frac{1}{n}[(1+\frac{x}{n})^{n-1}+...+1]\leq(1+\frac{x}{n})^{n-1}$$

For $n\rightarrow \infty$ we have

$$1\leq\frac{f(x)-1}{x}\leq f(x)$$.

For $x<0$ by substituting $x$ with $-x$ and dividing by $f(-x)>0$ we get:

$$\frac{1}{f(-x)}\leq \frac{\frac{1}{f(-x)}-1}{x}\leq 1$$

Now for all $x\neq 0$ we have:

$$\min\{1,e^x\}\leq\frac{e^x-1}{x}\leq\max\{1,e^x\}$$.

Using $\lim_{x\to 0} e^x=1$ and the sandwich rule we get: $$\lim_{x\to 0}\frac{e^x-1}{x}=1$$

I know I'm missing out on something important here but how did they get $(1+\frac{x}{n})^n$ from $e^x$ in the very first step?

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    Where is the proof from? – 2017-02-18
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    You might want to reread earlier chapters, it is very likely that this is how $e^x$ was defined. – 2017-02-18
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    @martycohen From my textbook, it's not written in English so I doubt that it would be helpful if I told you the title. – 2017-02-19
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    Nice and short proof by the way. +1 – 2017-02-19

2 Answers 2

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HINT: $$\lim_{n \to 0} \left(1 + \frac{x}{n}\right)^n = e^x$$

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Let $e^x-1=y$. Hence, $y\rightarrow0$ and since $\ln$ is a continuous function, we obtain: $$\lim\limits_{x\rightarrow0}\frac{e^x-1}{x}=\lim\limits_{y\rightarrow0}\frac{y}{\ln(1+y)}=\lim\limits_{y\rightarrow0}\frac{1}{\ln(1+y)^{\frac{1}{y}}}=1$$

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    This does not answer the question. – 2017-02-18