Lim 1 x 1 x e

We are going to show the following equality:

$$ \lim _{x \rightarrow 0}[1+x]^{\frac{1}{x}}=e $$

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

We have:

$$ \begin{aligned} \ln u[x]&=\ln [1+x]^{\frac{1}{x}}\\ &=\frac{1}{x} \ln [1+x]=\frac{\ln [1+x]}{x}\\ \end{aligned} $$

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 $\displaystyle\lim_{x \rightarrow c} f[x]=\lim _{x \rightarrow c} g[x]=0$ or $\pm \infty, g^{\prime}[x] \neq 0$ for all $x$ in $I$ with $x \neq c,$ and $\displaystyle\lim _{x \rightarrow c} \frac{f^{\prime}[x]}{g^{\prime}[x]}$ exists, then

$$ \lim _{x \rightarrow c} \frac{f[x]}{g[x]}=\lim _{x \rightarrow c} \frac{f^{\prime}[x]}{g^{\prime}[x]} $$

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

$$ \lim _{x \rightarrow 0} \frac{ln[1+x]}{x}=\lim _{x \rightarrow 0} \frac{\displaystyle\frac{1}{1+x}}{1}=1 $$

Second: using the definition of the derivative.

$$ \begin{aligned} \lim _{x \rightarrow 0} \frac{ln[1+x]}{x}&=\lim _{x \rightarrow 0} \frac{ln[1+x]-ln[1+0]}{x-0} \\ &=\lim _{x \rightarrow 0} \frac{f[x]-f[0]}{x-0}=f^{\prime}[0]=1 \end{aligned} $$
with $f[x]=\ln [1+x]$ and $\displaystyle f^{\prime}[x]=\frac{1}{1+x}$

We have:

$$ \begin{aligned} \lim _{x \rightarrow 0}[\ln u[x]]&=\lim _{x \rightarrow 0} \frac{\ln [1+x]}{x}=1\\ \exp[\lim _{x \rightarrow 0}[\ln u[x]]&=\exp [1]\\ \lim _{x \rightarrow 0}[\exp[\ln u[x]]&=e \\ \lim _{x \rightarrow 0}[u[x]]&=e \\ \end{aligned} $$

We conclude:

$$ \lim _{x \rightarrow 0}[1+x]^{\frac{1}{x}}=e $$

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