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

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.

## marked as duplicate by Guy Fsone, Parcly Taxel, muaddib, Chris Godsil, Arnaud D.Jan 28 '18 at 15:21

• If you don't have to use L'hopital, then have a look at this. – StubbornAtom Nov 13 '16 at 17:41
• In the last step, just split the thing you have on the left hand side. – 3x89g2 Nov 13 '16 at 17:42
• Easier to use $2\cdot 9^n> 4^n+9^n> 9^n$ and apply the squeeze theorem. – Thomas Andrews Nov 13 '16 at 17:42
• – Workaholic Nov 13 '16 at 18:22

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 $(4/9)^n+1$.
Divide top and bottom by $9^n$ to get:
$$\lim_{n\to\infty} \frac{(4/9)^n\ln 4+\ln 9}{(4/9)^n+1}$$
$$\lim_{n\to\infty}\frac{4^n\ln 4+9^n\ln 9}{4^n+9^n}=\lim_{n\to\infty}\frac{(4/9)^n\ln 4+\ln 9}{(4/9)^n+1}=\ln 9$$