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After studying limits on multivariable calculus, a question came to my mind.

For simplicity let $$L =\lim_{x\rightarrow 0} \frac{x^2}{x}$$

Before directly applying L'Hôpital's rule, one would directly "simplify" by division of $x$, resulting to $L=0$

Without consulting L'Hôpital's rule, or the definition for limits, why can we divide by $x$ if it also means "dividing by $0$" ?

I really need to clarify this since the multivariable counterpart is asking me to approach certain points through various directions.

Edit: I have used "dividing by 0" above with regards to the comment on this thread https://math.stackexchange.com/a/4190435/924242.

The user stated that multiplying the function $f(x)$ with another function $g(x) = 1$ and getting the limit of $f(x)g(x)$ at a point $a$ is not permitted if $g(x)$ is undefined at the said point

wd violet
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    You should go back and study what a limit means and is. Only if you have fully understood it, you're ready for multivariable calculus. – vitamin d Jul 05 '21 at 04:49
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    "why can we divide by $x$ if it also means dividing by $0$?" Dividing by $x$ here does not also mean dividing by $0$. The whole notion of a limit "$\lim_{x\to a}\cdots$" is predicated on ignoring what happens when $x$ is equal to $a$, and instead describing what happens as $x$ approaches $a$. (As I like to say, "Limits are about the journey, not the destination.".) So, you're free to divide-through by $x$ here, because, insofar as the limit context is concerned, $x$ is not $0$. – Blue Jul 05 '21 at 04:50
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    "if it also means dividing by $0$" — No, it does not mean that. The definition of the limit for $x \to x_0$ very specifically avoids evaluating the function at $x=x_0$. See for example Is it necessary that if a limit exists at a point it should be also defined at that point?. – dxiv Jul 05 '21 at 04:52
  • Thank you all for your clarifications. – wd violet Jul 05 '21 at 05:14
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    Just to be clear, I don't think that it's particularly strange that you're having this question now. In my experience and from what I've heard from other people, it's really pretty common to find little holes of knowledge like this when you start thinking more in-depth about a method you might have learned in one of your earlier classes, especially where you are now. So don't be too concerned if you find yourself asking a lot of questions like this, it's pretty normal. Instead just be glad that you're thinking about it enough to ask the questions – Stephen Donovan Jul 05 '21 at 06:09
  • When $x \rightarrow a$ , x is never equal to a. So you can cancel x-a. – Lawrence Mano Jul 05 '21 at 05:02

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Yes, that's the fun with limits. You can't divide with $0$, but as long as $x\to 0$ and not $x=0$ (that is, the magnitude of $x$ may be very very small, but it's not $0$), you can perform the division.

Note that you're only interested in the limiting behaviour, and not the value of the function at $x=0$ (which you clearly know, doesn't exist). And as long as the limit is what you want to deal with, you can perform that division with as small a quantity as you want.

Sayan Dutta
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