$ (1 + x)^{1/x} $ when $ x \to \infty$ without L’Hospital’s Rule Title is the question itself.
How can I show below without L’Hospital’s Rule
$$ \lim_{x\to \infty} ( 1 + x )^\frac{1}{x} = 1 $$
 A: For simplicity, let $f(h) = (1+1/h)^{h}$. We want to show $$\lim_{x\to\infty} f(x)=1$$
Or, $$\lim_{x\to\infty} \ln(f(x)) = \lim_{x\to\infty}\frac{\ln(1+x)}{x}=0.$$
Equivalently, for any $\epsilon>0$, then for $x$ large enough
$$1+x < e^{\epsilon x}\tag{1}$$
Intuitively, $(1)$ holds since $e^\epsilon>1$. More explicitly, if we let $e^\epsilon = 1+\delta$, then 
$$ e^{\epsilon x} = (1+\delta)^x  > 1+n\delta + \frac{n(n-1)}2\delta^2> 1+(n+1)>1+x$$
where $n=\lfloor x\rfloor$ and $x$ large enough.
Note: the above argument assumes that $x\to \color{red}{+}\infty$, similar argument works with $x\to\color{red}{-}\infty$, but with a little more effort.
A: For any $a>0$ and $x\ge 1$, we have $0\le \log(x)\le \frac{x^a-1}{a}$.
Hence, we can write for $x\ge 1$ and $0<a$
$$\begin{align}
\left(1+x\right)^{1/x}&=e^{\frac1x\log(1+x)}\\\\
&\le e^{\frac1x \frac{(1+x)^a-1}{a}}\tag 1
\end{align}$$
The inequality in $(1)$ is true for any $a>0$.  If we take $0<a<1$ (take $a=1/2$ for example), then $\lim_{x\to \infty}\left(\frac1x \frac{(1+x)^a-1}{a}\right)=0$.
Using the continuity of the exponential function, we conclude that 
$$\lim_{x\to \infty}\left(1+x\right)^{1/x}=1$$
A: For $(1+x)^{\frac{1}{x}}$ take the substitution $y=\frac{1}{x}$ where $y \rightarrow 0$ as $x \rightarrow + \infty$
Then the limit becomes:
$\lim_{x \rightarrow 0}(1+x)^{\frac{1}{x}}= \lim_{y \rightarrow 0}(1+\frac{1}{y})^y$
$$(1+\frac{1}{y})^y=e^{y\ln{(1+ \frac{1}{y})}}$$
Now find the limit  $$y \ln{(1+ \frac{1}{y})}=y\ln{(1+y)}-y \ln{(1-(1-y))}$$
using taylor expansion to:$$y\ln{(1+y)}$$ $$y \ln{(1-(1-y))}$$
at $y=0$
A: Based on Davit Mitra’s comment, I realized one simple proof.
(I am also very thankful to many other replies.)
By AM > GM,
$$ \frac{\overbrace{1 + 1 + \cdots + 1}^{x-2 \,\text{times}}+ \sqrt{1+x} + \sqrt{1+x}}{x} \ge (1 + x)^\frac1x \ge 1 \\
\underbrace{ 1 - \frac{2}x + \frac{2\sqrt{1+x}}x }_{(1)} \ge (1 + x)^\frac1x \ge 1$$
if $x$ goes to $\infty$, eq.(1) goes to $1$.  
A: HINT:
Let $x\ge 1$ and $n = [x]$ so $n\le x < n+1$. We have
$$\left(1+\frac{1}{n+1}\right)^n< (1+\frac{1}{x})^x < \left(1+\frac{1}{n}\right)^{n+1}$$
