$\lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}$. 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!

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.
