# Math Help - Limit_problem ass4_7

1. ## Limit_problem ass4_7

How I solve this ?

$lim_{x \to \infty} (x^{18}e^{-0.0001x})$

2. hi
$\lim_{x\to \infty }\frac{x}{8\exp(0.0001x)}=0$
hope it's right.

3. Originally Posted by gilyos
How I solve this ?

$lim_{x \to \infty} (x^18e^{-0.0001x})$
do you mean

$
x^{18}e^{-0.0001x}
$

$x^18e^{-0.0001x}$
$lim_{x \to \infty} (x^{18}e^{-0.0001x})$
5. This is an indeterminate form - $\frac{\infty}{\infty}$. You can use L'Hopital 18 times and you will be left with a scalar in the numerator, giving you the result that this limit is 0.
6. But you should recognize that any thing of the form $P(x)e^{-x}$, where P(x) is any polynomial, will go to 0 as x goes to infinity. $e^{-x}$ "dominates" any polynomial.