$\lim_{x\to\infty} x/e^x$ without L'Hospital's rule I rewrote the limit as $\lim_{x \rightarrow +\infty} x e^{-x}$ and so, applying the limit separately I would get $\infty\times 0$, which is indeterminate. 
How would I proceed from here?
 A: Firstly you prove that $$1+x<e^x$$ for all $x>0$. To prove this, let $f(x)=e^x-(1+x)$ and prove that its derivative is positive and therefore when $x>0$, $f(x)>f(0)$.
Next prove that $$1+x+\frac{x^2}{2}<e^x$$ for all $x>0$. The idea of the proof is similar and you will need to use your result in the first part.
Finally, after dividing both sides by $x$ (which is positive) from the result above, we have $$\frac{1}{x}+1+\frac{x}{2}<\frac{e^x}{x}.$$ When $x\to \infty$, the left hand side increases without bound, so does the right hand side. Thus your limit is zero.
The following is more than what you have asked, but using induction and the same idea, you can also prove that for all positive integers $n$,
$$\lim_{x\to \infty}\frac{e^x}{x^n}=\infty$$ and therefore,
$$\lim_{x\to \infty}\frac{x^n}{e^x}=0.$$
A: It would be easy to show if you are already aware that $e^x=1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+\cdots$
Now proceed as follows:
$\begin {align}
\lim_{x \to \infty}\frac x{e^x}&=\lim_{x \to \infty}\frac 1{e^x/x}\\
&=\lim_{x \to \infty}\frac 1{\bigg(1+x+\frac {x^2}{2!}+\frac{x^3}{3!}+\cdots\bigg)\bigg/x}\\
&=\lim_{x \to \infty}\frac 1{\frac 1{x}+1+\frac x{2!}+\frac {x^2}{3!}+\cdots}\\
&=\frac 1{0+1+\infty}\\
&=\frac 1{\infty}\\
&=0
\end {align}$
A: Let's consider the so defined succession:
$$x_1=\frac 1e  \ \ \ \ x_n=\frac{x_{n-1}}{e} \frac{n}{n-1}$$
It's easy to notice that:
$$x_n=\frac{n}{e^n}$$
The problem is reduced to:
$$L=\lim_{n\to \infty} x_n=\lim_{n\to \infty} \frac{x_{n-1}}{e} \frac{n}{n-1}$$
The degree of the numerator is the same of the denominator so:
$$L=\lim_{n\to \infty} \frac{x_{n-1}}{e}=\frac{1}{e}\lim_{n\to \infty} x_{n-1}=\frac Le$$
We have proved that:
$L=\frac Le \Rightarrow L=0 \ \ \ or  \ \ \ L=+\infty$
But since the sequence is strictly decreasing, in fact if for absurd:
$$x_n>x_{n-1}\Rightarrow \frac{n}{e(n-1)}>1 \Rightarrow \frac{n}{e(n-1)}>1 \Rightarrow n<\frac{e}{e-1} \ \ \ Absurd! $$
It follows that $L$ is finite and $L=0$
