Find $\displaystyle \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3}$. Apply l’Hopital’s rule as many times as necessary, verifying your results after each application

Printable View

- Dec 1st 2006, 04:58 PMFLTRl’Hopital’s rule
Find $\displaystyle \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3}$. Apply l’Hopital’s rule as many times as necessary, verifying your results after each application

- Dec 1st 2006, 05:33 PMtopsquark
$\displaystyle \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3}$

First both the numerator and denominator go to infinity in the limit, so we can apply L'Hopital's rule.

Thus:

$\displaystyle \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3} = \lim_{x\to\infty}\frac{2e^{2x}}{3(x+5)^2}$

Again, both numerator and denominator go to infinity so:

$\displaystyle \lim_{x\to\infty}\frac{2e^{2x}}{3(x+5)^2} = \lim_{x\to\infty}\frac{4e^{2x}}{6(x+5)}$

Again, both numerator and denominator go to infinity so:

$\displaystyle \lim_{x\to\infty}\frac{4e^{2x}}{6(x+5)} = \lim_{x\to\infty}\frac{8e^{2x}}{6}$

Finally, only the numerator goes to infinity and the denominator is a constant, so the net result is:

$\displaystyle \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3} = \lim_{x\to\infty}\frac{8e^{2x}}{6} \to + \infty$

or more precisely, the limit is undefined.

-Dan - Dec 1st 2006, 05:49 PMTriKri
Just another question, a little of topic maybe, how can one explain so I understand why l’Hopital’s rule is true when $\displaystyle f(c)=g(c)=\infty$? When $\displaystyle f(c)=g(c)=0$ it is easier.

- Dec 1st 2006, 05:57 PMtopsquark
- Dec 2nd 2006, 05:21 AMTD!