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 $-1leq sin (x) leq 1$ for all $x in mathbb{R}$ we have:

begin{align}

lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdot(-1)}{n+1}underbrace{leq}_{text{Is this fine?}}&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{(1)^{n}cdotsqrt{n}cdot(1)}{n+1}\ \ lim_{n rightarrow infty} frac{(-1)^{n+1}cdotsqrt{n}}{n+1}leq&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(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}cdotsqrt{frac{1}{n}}}{1+frac{1}{n}}leq&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(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}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq 0 \ \ therefore& lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(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!