# What is $\lim _{x\to \infty }\left(\frac{e^x}{x^n}\right)$?

What is $\lim _{x\to \infty }\left(\frac{e^x}{x^n}\right)$ ?

• It diverges to infinity. – Hanul Jeon Nov 12 '14 at 11:01

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 $+\infty$.
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.
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 $\infty$).
Another possibility: $$\lim_{x\to\infty}\left(\frac{e^x}{x^n}\right)= \lim_{x\to\infty}\left(\frac{e^{x/n}}{x}\right)^n= \left(\lim_{x\to\infty}\frac{e^{x/n}}{x}\right)^n=\cdots$$