# A seemingly simple application of L'Hopital's rule?

Here's the limit in consideration: $$\lim_{x\to -\infty} \frac{\ln(3^x+5^x)}{x}$$

I tried this limit with L'Hopital's Rule and seem to get a different answer from what WolframAlpha is saying which is $\ln(3)$. I wanted to do this without knowing some savvy manipulation of the numerator.

Attempt: Since we have the indeterminate form of $\frac{-\infty} {-\infty}$ we apply L'Hopital's rule giving us $$\frac{3^x(\ln3)+5^x(\ln5)} {3^x+5^x}$$ and then since we still have a indeterminate form, we apply L'Hopital's rule again, giving us $$\frac{3^x(\ln3)^2+5^x(\ln5)^2} {3^x(\ln3)+5^x(\ln5)}$$ which then we can factor out the common $3^x(\ln3)+5^x(\ln5)$ and then cancel resulting in simply $\ln(3)+\ln(5)$.

So, I got that $$\lim_{x\to -\infty} \frac{\ln(3^x+5^x)}{x}= \ln(3)+\ln(5)$$.

Where did I go wrong? Any help would be appreciated. Thanks!

• The mistake is we can factor out the common $3^x(\ln3)+5^x(\ln5)$ – Claude Leibovici Nov 29 '17 at 8:56

As Claude Leibovici wrote, your factorization is false. Since $3^x$ is falling slowlier than $5^x$ we have to factor out $3^x$ and get:
$$\frac{3^x\ln^2(3)+5^x\ln^2(5)}{3^x\ln(3)+5^x\ln(5)}=\frac{3^x\left(\ln^2(3)+\left(\frac53\right)^x\ln^2(5)\right)}{3^x\left(\ln(3)+\left(\frac53\right)^x\ln(5)\right)}=\frac{\ln^2(3)+\left(\frac53\right)^x\ln^2(5)}{\ln(3)+\left(\frac53\right)^x\ln(5)}$$ Since $\left(\frac53\right)^x\to 0$ for $x\to-\infty$ we get $$\lim_{x\to\infty}\frac{\ln(3^x+5^x)}x=\frac{\ln^2(3)}{\ln(3)}=\ln(3)$$ You can check what happens if you factor out $5^x$ instead. You get $$\frac{3^x\ln^2(3)+5^x\ln^2(5)}{3^x\ln(3)+5^x\ln(5)}=\frac{5^x\left(\ln^2(5)+\left(\frac35\right)^x\ln^2(3)\right)}{5^x\left(\ln(5)+\left(\frac35\right)^x\ln(3)\right)}=\frac{\ln^2(5)+\left(\frac35\right)^x\ln^2(3)}{\ln(5)+\left(\frac35\right)^x\ln(3)}$$ Since $\left(\frac35\right)^x\to\infty$ for $x\to-\infty$, you could use L'Hôpital again and you get \begin{align} \lim_{x\to\infty}\frac{\ln(3^x+5^x)}x&=\lim_{x\to-\infty}\frac{\ln^2(5)+\left(\frac35\right)^x\ln^2(3)}{\ln(5)+\left(\frac35\right)^x\ln(3)}=\lim_{x\to-\infty}\frac{\ln\left(\frac35\right)\left(\frac35\right)^x\ln^2(3)}{\ln\left(\frac35\right)\left(\frac35\right)^x\ln(3)} \\&=\lim_{x\to-\infty}\ln(3)=\ln(3) \end{align} So you get the same result. :-)
• Thank you for your detailed answer. A couple questions though. Can you explain how $3^x$ falling slowlier than $5^x$ results in only being able to factor out $3^x$ ?? Or more generally, how can you know that the factorization that I provided doesn't apply or when it does? – Jae Kim Nov 29 '17 at 18:41
• I'm not sure if I understand you correctly: I showed that if you can factor out $3^x$ or $5^x$ and both ways can be used to get the result. But $(3^x\ln(3)+5^x\ln(5))(\ln(3)+\ln(5))=3^x\ln^2(3)+3^x\ln(3)\ln(5)+5^x\ln(3)\ln(5)+5^x\ln^2(5)\neq 3^x\ln^2(3)+5^x\ln^2(5)$. – Mundron Schmidt Nov 29 '17 at 19:48
• And $3^x$ is falling slowlier than $5^x$ for $x\to-\infty$ because $\frac{5^x}{3^x}=\left(\frac53\right)^x\to 0$ for $x\to-\infty$ although $3^x$ and $5^x$ goes both to $0$ for $x\to-\infty$. – Mundron Schmidt Nov 29 '17 at 19:52
Consider also the following alternative solution without L'Hopital: write $$\frac{\log(3^x+5^x)}x=\log(3)+\frac{\log\left(1+(5/3)^x\right)}x$$ and note that the fraction on the right clearly tends to $0$.