find the limit: $\lim_{ n \to \infty }\sin ^2(\pi\sqrt{n^2+n})=?$ fine the limit:
$$\lim_{ n \to \infty }\sin ^2(\pi\sqrt{n^2+n})=?$$
my try :
$$\lim_{ n \to \infty }\sin ^2(\pi\sqrt{n^2+n})=\sin ^2(\lim_{ n \to \infty }\pi\sqrt{n^2+n})\\\sin^2(\infty)=$$Does not exist
is it right ?

$$\begin{align}\lim_{n\rightarrow \infty}\sin^2\left(\pi\sqrt{n^2+n}\right) &=\lim_{n\rightarrow \infty}\sin^2\left(n\pi-\pi\sqrt{n^2+n}\right)\\
&=\lim_{n\rightarrow \infty}\sin^2\bigg(\frac{-\pi^2 n}{n\pi+\pi\sqrt{n^2+n}}\bigg)\end{align}$$
So $$\lim_{n\rightarrow \infty}\sin^2\left(\pi\sqrt{n^2+n}\right) = 1$$
Because:$$\displaystyle \star \; \lim_{n\rightarrow \infty}\frac{\pi^2 n}{n\pi+\pi\sqrt{n^2+n}} = -\lim_{n\rightarrow \infty}\frac{\pi}{1+\sqrt{1+\frac{1}{n}}} = -\frac{\pi}{2}$$
