Calculating the limit $\lim \limits_{n \to \infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\frac{1}{n^2}) + 17}$ 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.
 A: $n>6:$
$$n^{\frac{1}{n}}\leq(4n+15)^{\frac{1}{n}}\leq \left(4n+\sin \sqrt{n}+\cos \frac{1}{n^2}\right)^{\frac{1}{n}}\leq (4n+19)^{\frac{1}{n}}\leq n^\frac{2}{n}$$
A: 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.
A: $\lim \limits_{n \to +\infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17}$
$4n+15 \le 4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17 \le 4n+19$
$\sqrt[n]{4n+15} \le \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17} \le \sqrt[n]{4n+19}$
$\lim \limits_{n \to +\infty} \sqrt[n]{4n+15} \le \lim \limits_{n \to +\infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17} \le \lim \limits_{n \to +\infty} \sqrt[n]{4n+19}$
A: $$
\begin{aligned}
\lim _{n\to \infty }\left(\left(4n\:+\:\sin \sqrt{n}\:+\:\cos \left(\frac{1}{n^2}\right)\:+\:17\right)^{\frac{1}{n}}\right)
& = \lim _{n\to \infty }\exp\left[\frac{\ln\left(4n\:+\:\sin \sqrt{n}\:+\:\cos \left(\frac{1}{n^2}\right)\:+\:17\right)}{n}\right]
\\& \approx \lim _{n\to \infty }\exp\left[\frac{\ln\left(4n\:\right)}{n}\right]
\\& = \color{red}{1}
\end{aligned}
$$
