Product of limit, sin, infinity, error? Hello I would like to know if there is a mistake :
I have to show that for any $t\geqslant0$ fixed
$$\lim_{n\to \infty}\sin\sqrt{t+4\pi n^{2}}=0$$
That's what I said,
Since $\sin(\cdot)$ is continuous and $$\sqrt{t+4\pi n^{2}}=2n\pi\sqrt{1+\frac{t}{4\pi n^{2}}}$$ for $n\geqslant1$.
$$\lim_{n\to \infty}\sin\sqrt{t+4\pi n^{2}}=\lim_{n\to \infty}\sin\left(2n\pi\sqrt{1+\frac{t}{4\pi n^{2}}}\right)$$
$$=\sin\left(\lim_{n\to\infty}2n\pi\sqrt{1+\frac{t}{4\pi n^{2}}}\right)=\\=\sin\left(\lim_{n\to\infty}2n\pi\cdot\lim_{n\to\infty}\sqrt{1+\frac{t}{4\pi n^{2}}}\right)=$$
$$=\sin\left(\lim_{n\to\infty}2n\pi\right)=\lim_{n\to\infty}\sin 2n\pi=\lim_{n\to\infty}0=0$$
 A: Assuming $n$ is real.
Let $t=4\pi^2 M>0$ fixed and $n=\sqrt{\pi(m^2-M^2)}$ varying then
$$\lim_{n\to \infty}\sin\sqrt{t+4\pi n^{2}}=\lim_{m\to \infty}\sin 2\pi m=0.$$
Let $t=4\pi^2 M>0$ fixed and $n=\sqrt{\pi((m+\frac14)^2-M^2)}$ varying then
$$\lim_{n\to\infty}\sin\sqrt{t+4\pi n^{2}}=\lim_{m\to \infty}\sin (2\pi m+\frac\pi 2)=1.$$
So the limit does not exist, if OP didn't change the expression.
A: Actually it does not exist the limit:$$\lim_{n\to \infty}\sin\sqrt{t+4\pi n^{2}}\;.$$
So, I think the OP intended to write the following limit:$$\lim_{n\to \infty}\sin\sqrt{t+4\pi^2n^{2}}\;.\quad(\text{ where }n\in\Bbb N\;)$$
Indeed ,
$\lim\limits_{n\to \infty}\sin\sqrt{t+4\pi^2n^{2}}=$
$=\lim\limits_{n\to \infty}\sin\left(\sqrt{t+4\pi^2n^{2}}-2\pi n\right)=$
$=\lim\limits_{n\to \infty}\sin\left[\dfrac{\left(\sqrt{t+4\pi^2n^{2}}-2\pi n\right)\left(\sqrt{t+4\pi^2n^{2}}+2\pi n\right)}{\sqrt{t+4\pi^2n^{2}}+2\pi n}\right]=$
$=\lim\limits_{n\to \infty}\sin\left(\!\dfrac t{\sqrt{t+4\pi^2n^{2}}+2\pi n}\!\right)=$
$=\sin\left(\!\lim\limits_{n\to \infty}\dfrac t{\sqrt{t+4\pi^2n^{2}}+2\pi n}\!\right)=$
$=\sin 0=0\,.$
Addendum:
I am going to prove that it does not exist the limit:$$\lim_{n\to \infty}\sin\sqrt{t+4\pi n^{2}}$$
without necessity of assuming that $\,n\,$ is any real number ( that is $\,n\,$ could be any positive integer ) .
If there existed the limit $\,\lim\limits_{n\to \infty}\sin\sqrt{t+4\pi n^{2}}=l\in\Bbb R\;,\;$ there would also exist the limit $\,\lim\limits_{n\to\infty}\sin\left(2\sqrt\pi n\right)=l\;,\;$ indeed
$\begin{align}
\lim\limits_{n\to\infty}&\,\sin\left(2\sqrt\pi n\right)=\\
&=\lim\limits_{n\to\infty}\sin\left(\sqrt{t+4\pi n^2}+2\sqrt\pi n-\sqrt{t+4\pi n^2}\right)=\\
&=\lim\limits_{n\to\infty}\left[\sin\left(\sqrt{t+4\pi n^2}\right)\cos\left(2\sqrt\pi n-\sqrt{t+4\pi n^2}\right)+\\
+\cos\left(\sqrt{t+4\pi n^2}\right)\sin\left(2\sqrt\pi n-\sqrt{t+4\pi n^2}\right)\right]=\\
&=\lim\limits_{n\to\infty}\left[\sin\left(\sqrt{t+4\pi n^2}\right)\cos\left(\!\!\dfrac{-t}{2\sqrt\pi n+\sqrt{t+4\pi n^2}}\!\!\right)+\\
+\underbrace{\cos\left(\sqrt{t+4\pi n^2}\right)}_{\text{it is bounded}}\;\underbrace{\sin\left(\!\!\dfrac{-t}{2\sqrt\pi n+\sqrt{t+4\pi n^2}}\!\!\right)}_{\text{it is an infinitesimal}}\right]=\\\\
&=l\cos0=l\;.
\end{align}$
Moreover,
$\begin{align}\lim\limits_{n\to\infty}&\cos\left(2\sqrt\pi n\right)=\\
&=\lim\limits_{n\to\infty}\dfrac{\sin\left[2\sqrt\pi(n+1)\right]-\sin\left[2\sqrt\pi(n-1)\right]}{2\sin\left(2\sqrt\pi\right)}=\\
&=\dfrac{l-l}{2\sin\left(2\sqrt\pi\right)}=0\,.
\end{align}$
On the other hand,
$0=\lim\limits_{n\to\infty}\cos\left(4\sqrt\pi n\right)=\lim\limits_{n\to\infty}\left[2\cos^2\left(2\sqrt\pi n\right)-1\right]=-1$
which is a contradiction.
Hence, there does not exist the limit $\,\lim\limits_{n\to\infty}\sin\sqrt{t+4\pi n^{2}}\,.$
