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I'm trying to find $\lim\limits_{x\to 0^+} \dfrac{x^2\sin(1/x)}{\sin x}$, I get $\dfrac{\infty}{0}$.

If $ \frac{\infty}{0} $is not an indeterminate form (like $ \infty \times 0, 1^\infty, \infty-\infty, \infty^0, 0^0 $) then what is it?

I know $ \frac{0}{\infty}$, for example, is zero...so what is $\frac{\infty}{0}$?

I'm asking because I'm trying to do $ \lim_{ x\to 0^+} [ (x^2)(\frac{\sin \frac{1}{x}}{\sin x }] $. When I plug in zero I get $ 0\times \frac{\infty}{0} $...so I'm not sure what to do after this. I tried separating the function like $\lim _{x\to 0^+}\frac{x^2}{\sin x}\times \lim_{x\to 0^+}\frac{\sin \frac{1}{x}}{\sin x} $....Then I got $\frac{0}{0}\times \frac{\infty}{0}$. For the $\frac{0}{0}$ one I applied L'hopitals and got 0. But then I still have that infinity/0. What do I do with this?

By the way, when I'm doing these "l'hopital's" problems...when do I know to stop or to keep going/rewriting. For example, if I get $ \frac{\infty}{0}$ do I just stop and assume that the limit does not exist...or do I try to rewrite and try to see if something works. Because I know that for those indeterminate forms I mentioned earlier, whenever I come across one of those, I know that I should just rewrite until I get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ to apply L'hopitals. But any time I get something other then these forms, I'm unsure of what to do.

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    $\begingroup$ First, you'd better look at your limit again, because it isn't $\frac{\infty}{0}$ In particular $\lim\limits_{x\to0^+} \sin\left (\frac{1}{x}\right ) \ne \infty$. $\endgroup$ – Paul Sinclair Sep 5 '15 at 4:52
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Observe:$$ \frac{x^2 \sin(1/x)}{\sin(x)}=\frac{x}{\sin(x)}\times \left[x \sin(1/x)\right] $$ The limit of the first term on the right is $1$ and the second is zero (these are both standard limits), so the limit you seek is $0$.

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While the question of how to handle your limit has been answered, I will attempt to answer your question of "What is $\frac{\infty}{0}$?".

Firstly, I'll clarify what is a "form" in this calculus context. A form like $\infty - \infty$ means: $\lim_{x} f(x)-g(x)$ where $\lim_{x} f(x)=\infty$ and similarly for $g$. In general, we talk about these things because we want to know to what extent the limits of the parts determine the limits of the whole.

Unambiguous forms: Things like $5-7$ (where one part has a limit of 5 and the other has a limit of 7) or $10/3$ are unambiguous forms (this is not a technical term), because of the limits of the parts exist and have those values, and the arithmetic makes sense and doesn't involve $\infty$, then theorems (often called "limit laws") guarantee that the limit of the whole is just what you get when you do the operation on the parts. There are also other unambiguous forms like $7*\infty$ (the limit of the whole is always $\infty$) or $0^\infty$ (the limit of the whole is always $0$) or $\frac{7}{\infty}$ (the limit of the whole is always $0$).

Indeterminate forms: The indeterminate forms you encounter are so called because they tell you essentially nothing about the limit of the whole. If you have a limit with the "form" $\infty-\infty$, then the limit of the whole might be $\infty$ or $-\infty$ or $17$ or "not exist due to oscillation" or whatever.

But there are other forms, which I will call "slightly ambiguous forms": like $1/0$, $\infty/0$, $0^{-\infty}$, etc. These forms are not indeterminate because if you take the limit of the absolute value of the whole function you always get $\infty$. For example, $\lim_{x\to\infty}\dfrac{1}{\frac{\sin x}{x}}$ has the form $1/0$, and the limit does not exist (due to "oscillation"), but $\lim_{x\to\infty}\left|\dfrac{1}{\frac{\sin x}{x}}\right|=\infty$. $\lim_{x\to\infty}\dfrac{x}{\frac{1}{x}}$ has the form $\frac{\infty}{0}$ but the limit simply equals $\infty$.

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