# Math Help - l’Hopital’s rule

1. ## l’Hopital’s rule

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

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

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

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

Again, both numerator and denominator go to infinity so:
$\lim_{x\to\infty}\frac{2e^{2x}}{3(x+5)^2} = \lim_{x\to\infty}\frac{4e^{2x}}{6(x+5)}$

Again, both numerator and denominator go to infinity so:
$\lim_{x\to\infty}\frac{4e^{2x}}{6(x+5)} = \lim_{x\to\infty}\frac{8e^{2x}}{6}$

Finally, only the numerator goes to infinity and the denominator is a constant, so the net result is:
$\lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3} = \lim_{x\to\infty}\frac{8e^{2x}}{6} \to + \infty$

or more precisely, the limit is undefined.

-Dan

3. Just another question, a little of topic maybe, how can one explain so I understand why l’Hopital’s rule is true when $f(c)=g(c)=\infty$? When $f(c)=g(c)=0$ it is easier.

4. Originally Posted by TriKri
Just another question, a little of topic maybe, how can one explain so I understand why l’Hopital’s rule is true when $f(c)=g(c)=\infty$? When $f(c)=g(c)=0$ it is easier.
All that my texts will say on the matter is that the proof is "technical." I assume that to be on the same level as "non-trivial" which is usually rather nasty.

-Dan

5. Originally Posted by TriKri
Just another question, a little of topic maybe, how can one explain so I understand why l’Hopital’s rule is true when $f(c)=g(c)=\infty$? When $f(c)=g(c)=0$ it is easier.
Proof.