Prove that $\frac{x}{e^x}$ tends to zero as $x \to \infty $ As the title states, I want to prove 
$$\lim_{x \to \infty} \frac{x}{e^x} =0$$
Clearly, L'Hopital's rule easily solves this. However, I'm curious to see if there's another way to prove it, without involving some differential or integral calculus (that is, by algebraic means). 
What I'm really interested about, is to prove that 
$$\lim_{x \to \infty} \frac{x}{e^{x^2}}=0 $$
I assume that proving the first limit will provide a way to prove the second one, using the squeeze method. If you know a direct way to prove the second limit, it will be more than perfect. 
Thanks in advance!
 A: For $x>0$, $e^x=1+x+x^2/2+\cdots>x^2/2$ so
$$0<\frac{x}{e^x}<\frac2x.$$
Likewise $e^{x^2}=1+x^2+\cdots>x^2$ so
$$0<\frac{x}{e^{x^2}}<\frac1x.$$
A: Mean Value Theorem
By the Mean Value Theorem, there is a $\xi\in(x/2,x)$ so that
$$
\begin{align}
\frac{e^x-e^{x/2}}{x-x/2}
&=e^{\xi}\\
&\ge e^{x/2}\tag1
\end{align}
$$
However, we also have
$$
\begin{align}
\frac{e^x-e^{x/2}}{x-x/2}\le2\,\frac{e^x}{x}
\end{align}\tag2
$$
Putting $(1)$ and $(2)$ together, we get
$$
\frac{x}{e^x}\le2e^{-x/2}\tag3
$$

Bernoulli's Inequality
Bernoulli's Inequality implies that
$$
\begin{align}
\left(1+\frac{x}n\right)^n
&\le\lim_{n\to\infty}\left(1+\frac{x}n\right)^n\\[6pt]
&=e^x\tag4
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{x}{e^x}
&\le\frac{x}{\left(1+\frac{x}2\right)^2}\\[3pt]
&\le\frac4x\tag5
\end{align}
$$
A: For the sequence by ratio test
$$a_n=\frac{n+1}{e^n}\implies \frac{a_{n+1}}{a_n}=\frac{n+2}{e^{n+1}}\frac{e^n}{n+1}=\frac{n+2}{n+1}\frac1e\to \frac1e<1\implies \frac{n+1}{e^n}\to 0$$
then observe that for $x\in (n,n+1)$
$$0<\frac{x}{e^x}\le\frac{n+1}{e^n} \to0$$
A: This is equivalent to
$\dfrac{\ln(x)}{x}
\to 0$
and this has been proved
many times here,
often by me.
One easy proof:
$\begin{array}\\
\ln(x)
&=\int_1^x \dfrac{dt}{t}\\
&<\int_1^x \dfrac{dt}{t^c}
\qquad\text{where } 0 < c < 1\\
&=\int_1^x t^{-c}dt\\
&=\dfrac{t^{1-c}}{1-c}|_1^x\\
&=\dfrac{x^{1-c}-1}{1-c}\\
&<\dfrac{x^{1-c}}{1-c}\\
\end{array}
$
Now choose $c$ appropriately.
For example,
$c=\frac12$ gives
$\ln(x)
\lt 2x^{1/2}$
so
$\dfrac{\ln(x)}{x}
\lt \dfrac{2}{x^{1/2}}
$.
Setting
$x = e^y$ gives
$\dfrac{y}{e^y}
\lt \dfrac{2}{e^{y/2}}
$.
If $c = 1-d$
where $d$ is small,
this is
$\ln(x)
\lt\dfrac{x^{d}}{d}
$
or
$\dfrac{\ln(x)}{x}
\lt \dfrac1{dx^{1-d}}
$.
In particular,
if $d = 1/n$,
this is
$\dfrac{\ln(x)}{x}
\lt \dfrac{n}{x^{1-1/n}}
$
or
$\dfrac{y}{e^y}
\lt \dfrac{n}{e^{y(1-1/n)}}
$.
A: First show $$\frac{x}{a^x}\to 0$$ for any $a>1$. This means 
$$\frac{x}{e^{2x}}\to 0$$ and write this as 
$$\frac{x^2}{e^{2x^2}}\to 0$$ now take the square root to get
$$\frac{x}{e^{x^2}}\to 0$$ 
