Product of limit, sin, infinity, error?

Question

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 $n\geqslant1$. $$\lim_{n\to \infty}\sin\sqrt{t+4\pi n^{2}}=\lim_{n\to \infty}\sin\left(2n\pi\sqrt{1+\frac{t}{4\pi n^{2}}}\right)$$ $$=\sin\left(\lim_{n\to\infty}2n\pi\sqrt{1+\frac{t}{4\pi n^{2}}}\right)=\\=\sin\left(\lim_{n\to\infty}2n\pi\cdot\lim_{n\to\infty}\sqrt{1+\frac{t}{4\pi n^{2}}}\right)=$$ $$=\sin\left(\lim_{n\to\infty}2n\pi\right)=\lim_{n\to\infty}\sin 2n\pi=\lim_{n\to\infty}0=0$$

Answer 1

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 $\,\lim\limits_{n\to\infty}\sin\sqrt{t+4\pi n^{2}}\,.$

Answer 2

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.