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I feel almost ashamed for putting this up here, but oh well..

I'm attempting to prove: $$\lim_{n\to \infty}\sqrt[n]{2^n+n^5}=2$$

My approach was to use the following inequality (which is quite easy to prove) and the squeeze theorem: $$\lim_{n\to \infty}\sqrt[n]{1}\leq \lim_{n\to \infty}\sqrt[n]{1+\frac{n^5}{2^n}}\leq \lim_{n\to \infty}\sqrt[n]{1+\frac{1}{n}}$$

I encountered a problem with the last limit though. While with functions using the $e^{lnx}$ "trick" would work, this problem is a sequences problem only, therefore there has to be a solution only using the first-semester-calculus-student-who's-just-done-sequences knowledge.

Or maybe this approach to the limit is overly complicated to begin with? Anyway, I'll be glad to receive any hints and answers, whether to the original problem or with $\lim_{n\to \infty}\sqrt[n]{1+\frac{1}{n}}$, thanks!

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Oh god, right, thanks for the edit, that's what I get for copying MathJax straight from the reference topic. – Dahn Jahn Dec 9 '12 at 11:25
You're welcome. And thanks for spending time to type mathematical expressions in mathematical markup language. – user1551 Dec 9 '12 at 11:55
up vote 1 down vote accepted

When $n$ is large $n^5<2^n$. So you can use: $$2<\sqrt[n]{2^n+n^5}<2^{1+1/n} $$

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Ah, that's very clever, thanks! – Dahn Jahn Dec 9 '12 at 11:42
Although this is not the highest rated answer (in fact, the only point is from me), because I find it elegant (what a subjective criterion..), I'm choosing it as the accepted answer. – Dahn Jahn Dec 9 '12 at 12:36

How about using $\root n\of{1+(1/n)}\le\sqrt{1+(1/n)}$ for $n\ge2$?

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Why not $1 < \sqrt[n]{1 + (1/n)} < 1 + (1/n)$? – Maikel Dec 9 '12 at 11:30
@manado --- yes, that's even better. – Gerry Myerson Dec 9 '12 at 11:56
Thanks. I suppose using a second squeeze theorem seemed too "out there" for my limited mathematical thinking :) – Dahn Jahn Dec 9 '12 at 12:12

The other answers provide easier approachs to your limit, that I would prefer. But the limit for your upper bound can be done straightforwardly: by definition, $$ \sqrt[n]{1+\frac1n}=\left(1+\frac1n\right)^{1/n}=e^{\frac1n\,\log(1+\frac1n)}. $$ As the exponential is continuous, all you need to do is to calculate the limit of $\frac1n\,\log(1+\frac1n)$, and this is trivial as it is a product where both factors go to zero.

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That's what I meant with not being able to use $e^{lnx}$, however, now when I see this answer, I am wondering if I was incorrect and maybe this really is a "fair" approach. It just seems a bit odd to use a function $e^x$ in a chapter where it's only being defined and not yet really used, if you know what I mean. Either way, thanks. – Dahn Jahn Dec 9 '12 at 12:29
Sorry, I missed that. My view is that making sense of $a^b$ without defining it as $e^{b\,\log a}$, is at best cumbersome. – Martin Argerami Dec 9 '12 at 12:34

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