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

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    $\begingroup$ It diverges to infinity. $\endgroup$ – Hanul Jeon Nov 12 '14 at 11:01
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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$.

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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.

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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 $\infty$).

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

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