# Value of $x^2 \sin (\frac1x)$ at $x=0$ [closed]

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 $y=0$?

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

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• $x^{2}$ 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 $x \to 0$. – jibounet Oct 9 '14 at 17:49
• Think about limits. – TZakrevskiy Oct 9 '14 at 17:49

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 !

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

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.

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 $x\rightarrow 0$.

• @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). – robjohn Oct 9 '14 at 19:16