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I have a question involving multi variable calculus:

$$\lim_{(x,y)\to(0,0)} \frac{x^4-4y^2}{x^2+2y^2}$$ When proving that limit exists... we sometimes let $y = mx$ to say that we are considering all straight line paths to the origin.

Then we write:

$$\lim_{x\to 0} \frac{x^4-4(mx)^2}{x^2+2(mx)^2}$$

I know some example of function but I want to understand the inner working of why the limit can be re-written as so?

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    This is a handy trick to prove that a limit doe not exist, but it will not prove that a limit does exist. To your question, though, for any limit, you can say $\lim_\limits{x\to a} f(x) = \lim_\limits {x\to b} f(g(x))$ if $g(x)$ is continuous and $g(b) = a$ – 2017-02-23
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    The change of (x, y) approaching (0, 0) to x approaching 0 is confusing me. – 2017-02-23

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In $\mathbb{R}^{2}$, we are not dealing with "one" direction anymore. Intuitively you may think of it by considering this question: In how many ways in a two-dimensional plane you can go to a given origin $(0,0)$? Apparently there are infinitely many crooked ways. So going to $(0,0)$ along a line is just one way out of infinitely many ways.

Note that the concept of multivariate limit just requires the existence regardless of in which way you approach $(0,0)$. Note that using a straight line we can reduce two variables to one. So it becomes easier to know if the limit does not exist. And yes, if you impose the relation $y=mx$ pertaining to $x$ and $y$, then logically of course you can write your first equation as the second one. However, as the definition of multivariate limit itself says, it is not sufficient to conclude a given multivariate limit exists if you just know it exists along every straight line.

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    I'm aware that I have to use precise definition of a limit to show that the limit does exist...but the limit notation in this situation is confusing me. Particularly which variables approach values. – 2017-02-23
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    @rert588. Oh okay. Actually I was answering according to what I thought you were asking. Perhaps you can try to describe your question more so that I would know if there is something I can help more :). – 2017-02-23
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    Is (x, y) --> (0, 0) still valid in the second equation and can I also use (x, mx) --> (0, 0)? – 2017-02-23
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    Why is x --> 0 valid in the second equation? – 2017-02-23
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    Yes because we now control the whole thing by $x$ alone. The reasoning is: If $y = mx$, then $(x,y) \to (0,0)$ if and only if $x \to 0$. – 2017-02-23
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    Thank you. I understand now. – 2017-02-23
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The limit cannot be re-written so. Looking at the lines $y=mx$ is a way to prove nonexistence of the limit, not existence of the limit. If the limit exists, then all lines give the same limit, but the converse is not true. If you check the lines and you get different limits depending on $m,$ you can stop and simply say the limit doesn't exist. If you check the lines and you get the same limit for each line, that's actually bad news: you're not done with the problem. You've then got to try another approach.