x sin(1/x) is continuous at 0

Example

Let $f(x) = x \sin(1/x)$ for $x \ne 0$, and let $f(0) = 0$. Then $f$ is continuous at $0$, and so $f$ is continuous on $\mathbb{R}$. Indeed, $|f(x)| \le |x|$ for all $x$, so $f(x) \to 0$ as $x \to 0$ by the squeeze theorem.

Remarks

The factor $x$ damps the oscillation of $\sin(1/x)$. The function is not differentiable at $0$. The difference quotient $(f(x) - f(0))/x = \sin(1/x)$ has no limit. The function $x^2 \sin(1/x)$ is differentiable with a discontinuous derivative.

Explanations

L1 human

The graph of $x \sin(1/x)$ oscillates infinitely often near $0$. But it stays between the lines $y = x$ and $y = -x$, which meet at the origin.