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 : )

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    $\begingroup$ 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. $\endgroup$
    – FShrike
    Oct 9, 2022 at 14:17
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    $\begingroup$ @311411 If the function is continuous for all real numbers, it must be continuous at x = 0 then. Should have made it clearer maybe. $\endgroup$ Oct 9, 2022 at 14:19
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    $\begingroup$ 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 $\endgroup$ Oct 9, 2022 at 14:19
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    $\begingroup$ @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 ? $\endgroup$ Oct 9, 2022 at 14:22
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    $\begingroup$ @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. $\endgroup$ Oct 9, 2022 at 14:38

1 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!


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