Prove that $x\sin(x)$ is continuous at $x = 0$

Question

Not too sure how to compute this. I can prove that $\sin(x)$ is continuous for all real numbers $x$ and hence $x=0$. I also know how to prove that $x$ is continuous for all real numbers $x$ and hence $x=0$. Then I could use the algebra of continuous functions product rule to show that this must mean $x\sin(x)$ is continuous at $x$ = 0.

There must be an easier way though right. My other thought:

I can prove using the sandwich theorem that |$x$||$\sin(x)$| = $0$ (or just trivially state it actually), so |$x$||$\sin(x)$| = $0$ < $\epsilon$, since $\epsilon$ > 0 as stated.

Cheers : )

Answer 1

The $\epsilon-\delta$ definition of continuity:

A function $f:\mathbb{R} \to \mathbb{R}$ is continuous at $x_0 \in \mathbb{R}$ if
for every $\epsilon>0$, there exists a $\delta>0$ such that
$0<|x-x_0|<\delta \implies |f(x)-f(x_0)| < \epsilon$

Consider this function, here, $f(x)=x\sin x$, $x_0=0$, and $f(x_0)=0$

Consider a given $\epsilon$. For that $\epsilon$, consider $\delta=\epsilon$
$0 < |x-x_0| < \delta \implies |x| < \delta $
Thus $|f(x)-f(x_0)| = |x \sin x - 0| \leq |x||\sin x| \leq |x| < \delta \implies |f(x)-f(x_0)| < \epsilon $

And we're done!