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This question already has an answer here:

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.

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ If you don't have to use L'hopital, then have a look at this. $\endgroup$ – StubbornAtom Nov 13 '16 at 17:41
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    $\begingroup$ In the last step, just split the thing you have on the left hand side. $\endgroup$ – 3x89g2 Nov 13 '16 at 17:42
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    $\begingroup$ Easier to use $2\cdot 9^n> 4^n+9^n> 9^n$ and apply the squeeze theorem. $\endgroup$ – Thomas Andrews Nov 13 '16 at 17:42
  • $\begingroup$ $\displaystyle\lim_{n\to\infty}\sqrt[n]{x^n+y^n}$. $\endgroup$ – Workaholic Nov 13 '16 at 18:22
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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 $(4/9)^n+1$.

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

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

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

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