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

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 : )

• Your first thought is correct and imo easy, you can’t really make it simpler than that. The second thought is wrong since continuity is about checking values of $x$ near to, but not at, zero. Oct 9, 2022 at 14:17
• @311411 If the function is continuous for all real numbers, it must be continuous at x = 0 then. Should have made it clearer maybe. Oct 9, 2022 at 14:19
• Have you used limits(for the question in the title)? The limits tend to zero, and it's zero atx=0, thus it's continuous Oct 9, 2022 at 14:19
• @mathandphysicsforever So you're saying that if I can prove the limit tends to zero as x tends to zero (using the squeeze theorem), then the function must be continuous at x = 0 using that theorem ? Oct 9, 2022 at 14:22
• @NikitaMazepin if you prove that the left and right limits are both 0, and also note that f(0)=0, then it is continuous by the definition of continuity. Oct 9, 2022 at 14:38

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!