# How do you find the limit e^x/x^3 as x->oo?

Dec 30, 2016

$\setminus {\lim}_{x \to \infty} \frac{{e}^{x}}{{x}^{3}} \to \infty$

#### Explanation:

This is in indeterminate $\frac{\infty}{\infty}$ form so you could plod through L'Hôpital's Rule 3 times.

I think it's more revealing to look at the exponential ${e}^{z}$ function which can be defined as

e^x =\sum _{k=0}^{\infty } x^{k} / (k!) =1+x+x^2 / 2+x^{3} / 6 + x^4 / 24 + O(x^5)

So we have

$\setminus {\lim}_{x \to \infty} \frac{1 + x + {x}^{2} / 2 + {x}^{3} / 6 + {x}^{4} / 24 + O \left({x}^{5}\right)}{{x}^{3}}$

$= \setminus {\lim}_{x \to \infty} \frac{1}{x} ^ 3 + \frac{1}{x} ^ 2 + \frac{1}{2 x} + \frac{1}{6} + \frac{x}{24} + O \left({x}^{2}\right)$

First three terms go to zero,

$\setminus {\lim}_{x \to \infty} \frac{1}{x} ^ 2 , \frac{1}{x} ^ 2 , \frac{1}{2 x} = 0$

the fourth is constant

$\setminus {\lim}_{x \to \infty} \frac{1}{6} = \frac{1}{6}$

... but look at the fifth and subsequent terms....

$= \setminus {\lim}_{x \to \infty} \frac{x}{24} + O \left({x}^{2}\right) \to \infty$