Proof of limit of lim (1+x)^(1/x)=e as x approaches 0

How to prove that limit of lim (1+x)^(1/x)=e as x approaches 0 ?

We are going to show the following equality:

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

Firt of all, we definie $u(x)=(1+x{)}^{\frac{1}{x}}$.

We have:

$$\begin{array}{rl}\ln u(x)& =\ln (1+x{)}^{\frac{1}{x}}\\ & =\frac{1}{x}\ln (1+x)=\frac{\ln (1+x)}{x}\end{array}$$

Two possibilities to find this limit.

First: L’Hôpital’s rule. L’Hôpital’s rule states that for functions $f$ and $g$ which are differentiable on an open interval $I$ except possibly at a point $c$ contained in $I$, if ${\displaystyle \underset{x\to c}{lim}f(x)=\underset{x\to c}{lim}g(x)=0}$ or $\pm \mathrm{\infty },g^{\prime }(x)\ne 0$ for all $x$ in $I$ with $x\ne c,$ and ${\displaystyle \underset{x\to c}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}}$ exists, then

$$\underset{x\to c}{lim}\frac{f(x)}{g(x)}=\underset{x\to c}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$$

Here $c=0$,$f(x)=\ln (1+x)$, $g(x)=x$. Which gives:

$$\underset{x\to 0}{lim}\frac{ln(1+x)}{x}=\underset{x\to 0}{lim}\frac{{\displaystyle \frac{1}{1+x}}}{1}=1$$

Second: using the definition of the derivative.

$$\begin{array}{rl}\underset{x\to 0}{lim}\frac{ln(1+x)}{x}& =\underset{x\to 0}{lim}\frac{ln(1+x)-ln(1+0)}{x-0}\\ & =\underset{x\to 0}{lim}\frac{f(x)-f(0)}{x-0}=f^{\prime }(0)=1\end{array}$$

with $f(x)=\ln (1+x)$ and ${\displaystyle f^{\prime }(x)=\frac{1}{1+x}}$

We have:

$$\begin{array}{rl}\underset{x\to 0}{lim}(\ln u(x))& =\underset{x\to 0}{lim}\frac{\ln (1+x)}{x}=1\\ \exp (\underset{x\to 0}{lim}(\ln u(x))& =\exp (1)\\ \underset{x\to 0}{lim}(\exp (\ln u(x))& =e\\ \underset{x\to 0}{lim}(u(x))& =e\end{array}$$

We conclude:

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