# Evaluate the limit of $\lim_{n\to\infty}\frac{n^{1/3}\sin(n!)}{1+n}$

I cannot seem to wrap my head around this problem due to its complexity. The answer seems to be that the limit goes to zero for the following reasons:

1. $\sin(x)$ will never be greater than $1$, and thus will only lessen the value of the numerator.
2. If $\sin(x)$ is omitted from the equation, the equation reads $\frac{n^{1/3}}{1+n}$. Therefore, because the power of the denominator is greater than that of the numerator, the denominator will reach infinity faster than the numerator, causing the limit to converge to zero.

This answer seems a bit trivial, but I cannot disprove the rationale. Can someone confirm or deny my thinking process? If denied, could you theorize a potential, more rigorous proof?

In fact we are using squeezing theorem. It states that

$$if\,\,\,\,\left\{ \matrix{ {l_n} \le {a_n} \le {u_n} \hfill \cr \mathop {\lim {l_n}}\limits_{n \to \infty } = 0 \hfill \cr \mathop {\lim {u_n}}\limits_{n \to \infty } = 0 \hfill \cr} \right.\,\,\,\,\,\,then\,\,\,\,\,\,\,\mathop {\lim {a_n}}\limits_{n \to \infty } = 0$$

Now consider the following

$$- {{{n^{{1 \over 3}}}} \over {1 + n}} \le {{{n^{{1 \over 3}}}\sin (n!)} \over {1 + n}} \le {{{n^{{1 \over 3}}}} \over {1 + n}}$$

since the lower and upper bound go to zero hence the sequence itself goes to zero. That's all. :)

Your reasoning is exactly the standard way of tackling that limit. Good job.

• That's great to hear, I've been struggling at analysis for a few weeks now, but I think I'm getting the hang of it – Kevin McDonough Oct 18 '15 at 18:19

Use the sandwich/squeeze theorem + algebra of limits + as you already mentioned the bounded property of the sine function if you want a more rigorous way to show the limit. Alternatively you could formally prove that it has the limit you suggest using the definition of a limit.

The shortest proof uses asymptotic analysis:

$$\frac{n^{1/3}}{1+n}\sim_\infty \frac{n^{1/3}}n=n^{-2/3}\xrightarrow[n\to\infty]{}0.$$ As $\,\lvert\sin(n!)\rvert\le 1$, $\;\dfrac{n^{1/3}\sin(n!)}{1+n}=O\bigl(n^{-2/3}\bigr)$, hence $\;\dfrac{n^{1/3}\sin(n!)}{1+n}$ tends to $0$ as $n$ tends to $\infty$.