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

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