Can you explain the behavior of $\frac{1}{x} \sin(\frac{1}{x})$ and $|\frac{1}{x} \sin(\frac{1}{x})|$ as $x$ approaches $0$

Question

When I graph it out, I can see what the behavior of the graph is, but my intuition is failing to grasp why it acts this way.

As $x\to 0$, $f(x) = \frac{1}{x} \sin(\frac{1}{x})$ only oscillates. If $\sin(\frac{1}{x})$ is bounded by $1$ and $-1$, and $\frac{1}{x}$ diverges to $\infty$, then why doesn't the whole function diverge? Shouldn't the divergence overwhelm the bounded sin function?

Then if we apply the absolute value, as $x\to 0$, $f(x) = |\frac{1}{x} \sin(\frac{1}{x})|$ now does diverge to $\infty$, but I don't understand what changed. $\sin$ now oscillates between $0$ and $1$, but somehow it is now overpowered by the divergence of $\frac{1}{x}$, where it wasn't before.

Answer 1

As to requested behaviour, then limit for both doesn't exist at $x \to 0$. Limit point 0 have both. Difference is that for $\left|\frac{1}{x} \sin(\frac{1}{x})\right|$ we have limit point $+ \infty$ and for $\frac{1}{x} \sin(\frac{1}{x})$ we have limit points $\pm \infty$

By the way, imho, main difference seems that $\left|\frac{1}{x} \sin(\frac{1}{x})\right|$ have no derivative at $x=\frac{1}{\pi n}$.

Answer 2

Both functions vanish along the sequence $(\frac 1 {n \pi})$ and are unbounded along the sequence $(\frac 1 {(n+\frac 1 2) \pi})$. Hence they both oscillate.