x sin x has no limit at infinity

Idea. The factor $x$ makes $f$ unbounded, which suggests $f\to+\infty$. But $\sin x$ keeps vanishing at the multiples of $\pi$. Two different sequences will show the contradiction.

First sequence. Let $x_n=2n\pi+\frac\pi2$. Then $x_n\to+\infty$ and $f(x_n)=x_n\cdot1=x_n\to+\infty$.

Second sequence. Let $y_n=n\pi$. Then $y_n\to+\infty$ and $f(y_n)=n\pi\cdot\sin(n\pi)=0$ for all $n$, so $f(y_n)\to0$.

Conclusion. If $f$ had a limit (finite or infinite) at $+\infty$, it would be the limit along every sequence tending to $+\infty$. But the two sequences give $+\infty$ and $0$. So $f$ has no limit at $+\infty$.

Remark. The function is unbounded, yet it does not tend to infinity. This is a good example showing that "unbounded" and "tends to infinity" are different properties.