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I am trying to calculate this limit:

$$ \lim \limits_{n \to \infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17}$$

I understand I should use squeeze theorem but I am having some trouble applying it to this particular formula.

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Do you know how to compute the limit for $n^{1/n}$? – Marek Jan 26 '13 at 15:24
up vote 1 down vote accepted

$\lim \limits_{n \to +\infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17}$

$4n+15 \le 4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17 \le 4n+19$

$\sqrt[n]{4n+15} \le \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17} \le \sqrt[n]{4n+19}$

$\lim \limits_{n \to +\infty} \sqrt[n]{4n+15} \le \lim \limits_{n \to +\infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17} \le \lim \limits_{n \to +\infty} \sqrt[n]{4n+19}$

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$$n^{\frac{1}{n}}\leq(4n+15)^{\frac{1}{n}}\leq \left(4n+\sin \sqrt{n}+\cos \frac{1}{n^2}\right)^{\frac{1}{n}}\leq (4n+19)^{\frac{1}{n}}\leq n^\frac{2}{n}$$

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This limit goes to 1. If you have any polynomial $P$ then $$\lim_{n\to\infty} |P(n)|^{1/n} = 1.$$ Since your radicand is bounded by a polynomials above and below, your limit is 1.

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