# L'Hopital's rule hw help

#1

I am finding that L’Hopital’s rule does not work for this problem and that the answer is simply infinity. But apparently that is not correct. Anyone know what to do?

#2

There’s a couple of ways to do this. If you want to coerce into a form where l’Hospital’s rule is applicable, you can do the following:

\left(\frac{15x}{15x+1}\right)^{9x} = e^{\ln\left(\frac{15x}{15x+1}\right)^{9x}} = e^{9x\ln\left(\frac{15x}{15x+1}\right)}.

We can now focus on just the exponent and write:

\begin{align} 9x\ln\left(\frac{15x}{15x+1}\right) &= \frac{9\left(\ln(15x) - \ln(15x+1)\right)}{1/x} \\ &\sim \frac{9\left(15/(15x) - 15/(15x+1)\right)}{-1/x^2} \\ &= -\frac{9 x^2}{x (1+15 x)} = -\frac{9 x}{1+15 x} \to -\frac{9}{15}=-\frac{3}{5} \end{align}

Thus, the answer is e^{-3/5} = 1/e^{3/5}.

#3

I am not finding this to work either.
I followed it exactly!
Pointing out that a different number was used in the solving, I ask if that would change anything?
also…
What did you do with the x’s at the end?
How did - 9/ 1+15x turn into just -9/15??

#4

We take a limit as x\to\infty.

Well it’s not just -9/1+15x, it’s -9x/(1+15x) and the limit as x\to\infty of that is -9/15.

Yes, if the numbers in the question change, I’m sure the answer would change. The technique should be the same, though.