What is $\lim _{x\to \infty }\left(\frac{e^x}{x^n}\right)$? What is $\lim _{x\to \infty }\left(\frac{e^x}{x^n}\right)$
?
 A: Recall that 
$$
e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=1+\frac{x}{1!}+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}+\ldots
$$
then, for each fixed $n$, as $x \to +\infty$, you get
$$
e^x>\frac{x^{n+1}}{(n+1)!}
$$ and
$$
\frac{e^x}{x^n}>\frac{x}{(n+1)!} \longrightarrow +\infty, \quad\text{as} \quad x \to +\infty,
$$ and the desired limit is $ +\infty$.
A: The limit is $+\infty$.
One way to see that is to use L'Hopital's rule. Differentiate numerator and denominator $n$ times, and you get
$$\frac{e^x}{n!}$$
which clearly diverges to positive infinity.
A: Exponential function increases faster than any polynomial function. You can imagine it this way - an exponential function consists of polynomials of progressively higher and higher degrees summed together. For any degree $n$ polynomial you are comparing against, exponential will have terms of degree $n+1$ and beyond too. (Agreed with progressively smaller coeffs, but eventually you are taking the limit $x \rightarrow \infty$, so that's hardly helping!)
As $x \rightarrow \infty$, the numerator increases much faster than the denominator, and the expression diverges (tends to $\infty$). 
A: Another possibility:
$$\lim_{x\to\infty}\left(\frac{e^x}{x^n}\right)=
\lim_{x\to\infty}\left(\frac{e^{x/n}}{x}\right)^n=
\left(\lim_{x\to\infty}\frac{e^{x/n}}{x}\right)^n=\cdots$$
