1
$\begingroup$

I am having a problem calculating the following limit:

$\lim\limits_{n \to \infty} \frac{n^{2/3} \sin(n!)}{n+1}$

Please help.

Thank you in advance

  • 0
    **Hint:** $|\sin x|\le 1$.2012-11-14

2 Answers 2

2

Think about what's happening as $n\to \infty$: $n^{2/3}$ and $n$ are both growing without bound, but $\sin n!$ is bounded above by $1$, since the sine function can only take on values between $-1$ and $1$. So as $n$ increases, the $\sin n!$ term becomes inconsequentially small compared to the others. Then

$\displaystyle \lim_{n\to\infty} \frac{n^{2/3}\sin(n!)}{n+1} = \lim_{n\to\infty} \frac{n^{2/3}}{n+1}$

Now since $n^{2/3}$ is a smaller power than $n$, it grows more slowly towards infinity. Thus

$ \lim_{n\to\infty} \frac{n^{2/3}}{n+1}= 0.$

So the limit of the sequence is $0$.

3

Since

$0\leq\frac{n^{2/3}}{n+1}\leq\frac{n^{2/3}}{n}=\frac{1}{\sqrt[3]n}\xrightarrow[n\to\infty]{} 0$

and also $\,|\sin (n!)|\leq 1\,$ ,as you can use the following easy-to-prove lemma:

Lemma: If $\,a_n\xrightarrow [n\to\infty]{} 0\,$ and $\,\{b_n\}\,$ bounded, then $a_nb_n\xrightarrow [n\to\infty]{} 0\,$