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What is the value of $y=x^2 \sin(1/x)$ at $x=0$?


I see that $x^2 =0$ but $\sin (1/x)$ is undefined.

More generally: if a function made up of a product of functions, like $y= a(x)b(x)c(x)d(x)\dots$, then at a specific value of $x$ if one of the functions is $0$ and all the other functions are undefined, does it still mean that $y=0$?

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closed as unclear what you're asking by daw, Lost1, Hamou, apnorton, Hakim Oct 9 '14 at 21:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ $x^{2}$ goes to $0$ as $x \to 0$ and $x \, \mapsto \, \sin \Big(\frac{1}{x}\Big)$ is bounded. So $x^{2} \sin \Big( \frac{1}{x} \Big) \, \to \, 0$ as $x \to 0$. $\endgroup$ – jibounet Oct 9 '14 at 17:49
  • $\begingroup$ Think about limits. $\endgroup$ – TZakrevskiy Oct 9 '14 at 17:49
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The function $x \, \longmapsto \, x^{2} \sin \Big( \frac{1}{x} \Big)$ is not defined at $x = 0$ because "$1/0$" does not exist. Note also that this function is continuous on $\mathbb{R}^{\ast}$.

However, this is not all we can say. Since the function $x \, \longmapsto \, \sin \Big( \frac{1}{x} \Big)$ is bounded, one can prove that :

$$ \lim \limits_{x \to 0} x^{2} \sin \Big( \frac{1}{x} \Big) = 0.$$

As a consequence, even though the function $x \, \longmapsto \, x^{2} \sin \Big( \frac{1}{x} \Big)$ is not defined at $x=0$, it has a finite limit as $x$ goes to $0$. So, this function can be extended to a new function, say $\overline{f}$, defined on $\mathbb{R}$ the following way :

$$ \overline{f}(x) = \begin{cases} x^{2} \sin \Big( \frac{1}{x} \Big) & \text{if } x \neq 0 \\[2mm] 0 & \text{if } x = 0 \end{cases} $$

This new function $\overline{f}$ is defined at $x=0$ and continuous on $\mathbb{R}$. In a very unformal, non-mathematical way, we could say that the value of the function $x \, \longmapsto \, x^{2} \sin \Big( \frac{1}{x} \Big)$ at $x=0$ is $0$. But this is not really the truth !

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If $f(x) = x^2\sin (1/x)$, then $f$ is undefined at $x = 0$, but $\lim_{x\to 0} f(x) = 0$.

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When $x=0$ , then $y= 0\sin(\frac10)$ as $\sin(\frac10)=$undefined then $y=$ undefined as the whole equation becomes undefined when at least one term is.

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Pretty sure if one of the functions is undefined $y$ must also be undefined.

Pretty sure this is a question about limits though where $y\rightarrow 0$ when $x\rightarrow 0$.

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  • $\begingroup$ @TheGame: The first sentence does answer the question, although in a rather non-committal way. The second sentence is a suggestion about how the function should be defined at $0$ for it to be continuous (which may or may not have been a part of the context behind the question). $\endgroup$ – robjohn Oct 9 '14 at 19:16

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