Question

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

Answer 1

`\lim_(x to oo) (e^x)/(x^3) to oo`

Explanation:

This is in indeterminate `oo/oo` 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

`\lim_(x to oo) (1+x+x^2 / 2+x^{3} / 6 + x^4 / 24 + O(x^5))/(x^3)`

`= \lim_(x to oo) 1/x^3+1/x^2+1 /( 2x) + 1 / 6 + x / 24 + O(x^2)`

First three terms go to zero,

`\lim_(x to oo) 1/x^2, 1/x^2, 1 /( 2x) = 0`

the fourth is constant

`\lim_(x to oo) 1 / 6 = 1/6`

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

`= \lim_(x to oo) x / 24 + O(x^2) to oo`