# How can I find this limit? L'hopital's rule is never-ending

I have the following limit $$\lim_{x\to\infty}\frac{-7^x}{(\ln11-\ln7)\cdot11^x}$$ Graphing gives me $$0$$ but plugging it in gives infinity over infinity.

I then tried L'Hôpital's rule and it seems to fall into a never ending loop. I can't seem to simplify this - what would be the best way to find the limit?

Edit: would like to clarify I want to know how to find it algebraically

• Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways. – egreg Jan 12 '19 at 23:53
• @egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment) – Andrew Wang Jan 13 '19 at 2:25

You can write the expression as $$-\left(\ln \frac{11}{7}\right)^{-1} \left(\frac{1}{(11/7)}\right)^x$$, which goes to zero as $$x \rightarrow \infty$$
Rewrite $$\frac{-7^x}{(\ln11-\ln7)\,11^x}=- \frac{\exp(x \log(7))}{\log \left(\frac{11}{7}\right) \exp(x \log(11))}=-\frac{\exp(x \log(7))\,\exp(-x \log(11))}{\log \left(\frac{11}{7}\right)}=-\frac{\exp(-x (\log(11)-\log(7)))}{\log \left(\frac{11}{7}\right)}=-\frac{\exp(-x \log \left(\frac{11}{7}\right))}{\log \left(\frac{11}{7}\right)}$$ and take ito account the fact that $$\log \left(\frac{11}{7}\right)>0$$