Prove $\lim\limits_{x \to \infty} x \sin^2 x$ does not exist.

Question

I need to prove $\lim\limits_{x \to \infty} x \sin^2 x$ does not exist, and request your help verifying my argument is correct. I've got somewhat decent at making $\epsilon-\delta$ arguments, but proving a limit does not exist has been a bit challenging actually.

We need to prove $\lim\limits_{x \to \infty} x \sin^2 x \neq \ell$, for every $\ell \in \mathbb{R}$, so take $\ell \in \mathbb{R}$. The result follows if we can find some $\epsilon > 0$ such that for every $N \in \mathbb{R}$, there is some real $x$ such that $x > N$ but $\lvert x \sin^2 x - \ell \rvert \geq \epsilon$.

If $\epsilon =\lvert \ell \rvert + 1$, then for every $N \in \mathbb{R}$ we can find $x_0 > \max(N, \quad 2\lvert \ell \rvert + 1)$ such that $x_0 = \cfrac{\pi}{2} + 2\pi k$, for some big enough $k \in \mathbb{Z}$. For this $x_0$ we have that $$x_0 \sin^2 x_0 = x_0 \cdot 1 = x_0 > 2\lvert \ell \rvert + 1 \geq 1 + \lvert \ell \rvert + \ell$$. This implies $x_0 \sin^2 x_0 - \ell = x_0 - \ell > 1 + \lvert \ell \rvert = \epsilon$ and the proof is complete.

Any comments or suggestions for improvement are very appreciated.

Answer 1

Your argument is fine, though as mentioned in the comments you don't address the possibility that the limit is infinite.

Here is a slightly simpler idea. Given any $k>0$, then $x_k=\frac\pi2+2k\pi>k$ and $x_k\sin^2x_k=x_k>k$. Also $z_k=2k\pi>k$ and $z_k\sin^2z_k=0$. So $$ \lim_{k\to\infty}x_k\sin^2x_k=\infty, $$ while $$ \lim_{k\to\infty}z_k\sin^2z_k=0. $$ This precludes the existence of the limit (finite or infinite), since if it existed all sequences that go to infinity would give the same limit.

Answer 2

This is just a variation of Martin Argerami's answer. He constructed a sequence $(x_k)$ such that $\lim_{k\to\infty}x_k\sin^2x_k=\infty$. Let us show that for each $a \in [0,\infty)$ there exists a sequence $(y_k)$ such that $$\lim_{k\to\infty}y_k\sin^2y_k=a .$$

We have $\sin^2 x = 0$ for $x = m\pi$ and $\sin^2 x = 1$ for $x = \pi/2 + m\pi$ for all $m \in \mathbb Z$. Thus by the IVT the function $f(x) = x \sin^2x$ maps the interval $[m\pi, \pi/2 + m\pi]$ onto the interval $[0,\pi/2 + m\pi]$. Hence for $m \ge m_0 = \max(0,(a-2)/\pi)$ we find a point $y_m \in [m\pi, \pi/2 + m\pi]$ such that $f(y_m) = a$. Let $y_m = m\pi$ for $m < m_0$. Then clearly $y_m \to \infty$ as $m \to \infty$ and $\lim_{m\to\infty}y_m\sin^2y_m=a$.