Prove $\lim\limits_{x \to \infty} x \sin^2 x$ does not exist. 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 $\epsilon &gt; 0$ such that for every $N \in \mathbb{R}$, there is some real $x$ such that $x &gt; N$ 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 $x_0 &gt; \max(N, \quad 2\lvert \ell \rvert + 1)$ 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
$$x_0 \sin^2 x_0 = x_0 \cdot 1 = x_0 &gt; 2\lvert \ell \rvert + 1 \geq 1 + \lvert \ell \rvert + \ell$$.
This implies $x_0 \sin^2 x_0 - \ell = x_0 - \ell &gt; 1 + \lvert \ell \rvert = \epsilon$ and the proof is complete.
Any comments or suggestions for improvement are very appreciated.
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 $k&gt;0$, then $x_k=\frac\pi2+2k\pi&gt;k$ and $x_k\sin^2x_k=x_k&gt;k$. Also $z_k=2k\pi&gt;k$ 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.
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 $m &lt; m_0$. Then clearly $y_m \to \infty$ as $m \to  \infty$ and $\lim_{m\to\infty}y_m\sin^2y_m=a$.