# Math Help - L'Hopital's Rule

1. ## L'Hopital's Rule

I need to apply L'hopital's rule as x approaches infinity for the following:

(e^2x)/((x+5)^3)

which gives me:

(2e^2x)/((x+5)^3) - (3e^2x)/((x+5)^4)

and a second time gives me:

(4e^2x)/((x+5)^3) - (12e^2x)/((X+5)^4) + (12e^2x)/((X+5)^5)

I guess my question is, how do I know when to stop (i.e. when have I found the limit as x approaches infinity with L'Hopital's rule?)

2. Apply L'hôpital's Rule until you get something except the indeterminate form. You should also stop when you get $\mp \infty$ or $0$ too.

3. Hello, pennydooodle!

You don't seem to have a grip on L'Hopital's Rule . . .

Apply L'Hopital's Rule: . $\lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3}$
You seem to be using some kind of strange Quotient Rule here . . .

Differentiate the numerator: . $2e^{2x}$

Differentiate the denominator: . $3(x+5)^2$

. . Then: . $\;\lim_{x\to\infty}\frac{2e^{2x}}{3(x+5)^2} \quad\Rightarrow\quad \frac{\infty}{\infty}$

Apply L'Hopital again.

$\begin{array}{cc}\text{Numerator:} & 4e^{2x} \\
\text{Denominator:} & 6(x+5) \end{array}$

. . Then: . $\lim_{x\to\infty}\frac{4e^{2x}}{6(x+5)} \quad\Rightarrow\quad \frac{\infty}{\infty}$

Apply L'Hopital again:

$\begin{array}{cc}\text{Numerator:} & 8e^{2x} \\
\text{Denominator: } & 6 \end{array}$

. . Then: . $\lim_{x\to\infty}\frac{8e^{2x}}{6} \quad\Rightarrow\quad \infty$

4. Thank you Soroban!

Now I see what I was doing wrong. I was attempting to differentiate the entire equation rather than each part on its own.

Thank you again!