Finding the limit $\lim_{n\to\infty}\sqrt[n]{4^n+9^n}$ using L'Hopital's rule

Question

The problem was to find

$$\lim_{n\to\infty}\sqrt[n]{4^n+9^n}$$

So after a couple tries what I did was to take the natural logarithm of the limit so

$$\lim_{n\to\infty}\sqrt[n]{4^n+9^n}=L$$ $$\downarrow$$ $$\lim_{n\to\infty}\ln(\sqrt[n]{4^n+9^n})=\ln L$$ $$\downarrow$$ $$\lim_{n\to\infty}\frac{\ln({4^n+9^n})}{n}=\ln L$$ $$\downarrow L'Hopital$$ $$\lim_{n\to\infty}\frac{4^n\ln 4+9^n\ln 9}{4^n+9^n}=\ln L$$

And there I'm stuck. I checked in Wolfram and $\lim_{n\to\infty}$ of both the initial function and the one after L'Hopital's rule is $9$. ($\ln L=\ln 9\rightarrow L=9$).

I'd like to know how to find the limit from the last step I made, and if there's a more elegant way of solving the problem (which I'm sure there is), maybe without using L'Hôpital's rule.

Thanks.

Answer 1

Divide top and bottom by $9^n$ to get:

$$\lim_{n\to\infty} \frac{(4/9)^n\ln 4+\ln 9}{(4/9)^n+1}$$

Answer 2

L'Hospital is a detour.

If you put a factor of $9$ outside the root sign, we get $$ \sqrt[n]{4^n+9^n} = 9 \cdot \sqrt[n]{(4/9)^n+1} $$

Here $(4/9)^n$ goes to $0$, and taking an $n$th root yields something even closer to $1$ than $(4/9)^n+1$.

Answer 3

$$\lim_{n\to\infty}\frac{4^n\ln 4+9^n\ln 9}{4^n+9^n}=\lim_{n\to\infty}\frac{(4/9)^n\ln 4+\ln 9}{(4/9)^n+1}=\ln 9$$