$\lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}$. Am I correct? I have to find this limit:

\begin{align} \lim_{n \rightarrow \infty}
 \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1} \end{align}

My attempt:
Since we know that $-1\leq \sin (x) \leq 1$ for all $x \in \mathbb{R}$ we have:
\begin{align}
\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot(-1)}{n+1}\underbrace{\leq}_{\text{Is this fine?}}&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{(1)^{n}\cdot\sqrt{n}\cdot(1)}{n+1}\\ \\ \lim_{n \rightarrow \infty} \frac{(-1)^{n+1}\cdot\sqrt{n}}{n+1}\leq&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{n+1}
\end{align}
My doubt in the inequeality is because of the $(-1)$ terms. If everything is correct, then we have
\begin{align}
\lim_{n \rightarrow \infty} \frac{(-1)^{n+1}\cdot\sqrt{\frac{1}{n}}}{1+\frac{1}{n}}\leq&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}}}{1+\frac{1}{n}}\\ \\  \Rightarrow \ \ \ 0 \leq &\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq 0 \\ \\ \therefore& \lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1}=0
\end{align}
Am I correct? If I am not, how can I solve it? There are other ways to find this limit? I really appreciate your help!
 A: The first inequality is not true in general: if $a\leqslant b$ and $c<0$, then $ac\color{red}{\geqslant}bc$.
But note that$$\left|\frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}\right|\leqslant\frac{\sqrt n}{n+1}\to0,$$and therefore your limit is $0$ indeed.
A: 
If $(a_n)$ is a bounded sequence and $(b_n)$ is an infinitesimal sequence, then the product sequence $(a_n \cdot b_n)$ is also infinitesimal.

Starting from your sequence,
$$\lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}$$
Being $\forall n\in \mathbb{N}$ then $-1 \leq (-1)^{n} \le 1$
and $-1\leq \sin (n^n)\leq 1$  two bounded sequences then $$\lim_{n \rightarrow \infty} \frac{(-1)^{n}}{n+1}=\lim_{n \rightarrow \infty} \frac{\sin (n^n)}{n+1}=0$$
$$\lim_{n \rightarrow \infty} \underbrace{[(-1)^{n}\sin(n^{n})]}_{\text{bounded sequences}}\,\frac{\sqrt{n}}{n+1}=\lim_{n \rightarrow \infty} [(-1)^{n}\sin(n^{n})]\cdot \underbrace{\sqrt{\frac{n}{(n+1)^2}}}_{\text{infinitesimal sequence}}=0$$
A: 
\begin{align} \lim_{n \rightarrow \infty}
 \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1} \end{align}

This is one of those problems where you are supposed to take a step back and make things easy on yourself.
Let
$$f(n) = (-1)^{n}\sin(n^{n})$$
and let
$$g(n) = \frac{\sqrt{n}}{n+1} < \frac{1}{\sqrt{n}}.$$
You want
$$\lim_{n\to \infty} [f(n) \times g(n)].$$
As $n \to \infty, f(n)$ is a bounded function. 
That is $~-1 \leq f(n) \leq +1~$ for all $n$.
Also, clearly, as $n \to \infty, g(n) \to 0.$
Therefore, as $n \to \infty, \{f(n) \times g(n)\} \to 0.$
Edit
Adding some formality:
For all $n, |f(n) \times g(n)| \leq |g(n)|$ which goes go zero.
Therefore, since $|f(n) \times g(n)| \to 0,~ \{f(n) \times g(n)\} \to 0.$
A: Fact 1.  If $a_n\leq b_n$ for large $n$ then $\displaystyle\lim_{n\to\infty}a_n\leq\lim_{n\to\infty}b_n$, provided the limits exist.
Fact 2.  If $\displaystyle\lim_{n\to\infty}|a_n|=0$ then $\displaystyle\lim_{n\to\infty}a_n=0$.
From fact 1 we have
$$
0\leq\lim_{n\to\infty}\left|\frac{(-1)^n\sqrt{n}\sin(n^n)}{n+1}\right|
\leq\lim_{n\to\infty}\frac{\sqrt{n}}{n+1}
=0
$$
and so we can apply fact 2 to obtain
$$
\lim_{n\to\infty}\frac{(-1)^n\sqrt{n}\sin(n^n)}{n+1}
=0.
$$
A: You're either right or very nearly right: in either case I would be more comfortable putting some absolute value signs on the function to get rid of the $(-1)^n$ terms.
Here's why: let's say we have some $f(x)$ such that $\lim_{x \to a} |f(x)| = 0$. Notice by the linearity of limits that $\lim_{x \to a} -|f(x)| = -\lim_{x \to a} |f(x)| = 0$.
So, because for any $f(x)$ we have $-|f(x)| \leq f(x) \leq |f(x)|$, by the squeeze theorem we have that $\lim_{x \to a} f(x) = 0$.
So, the $(-1)^n$ terms aren't an issue, but I would state this explicitly in your work.
