I'm having hard time understanding what is the correct way to prove whether the limit exists or not for multivariable functions.

For example, let's say we have a function
$$\lim_{(x,y)\rightarrow(0,0)}\frac{x-y}{x+y} $$

Why can't I just plug in the values, 0 and 0 and show that this limit DNE?
The solution that I have is the following:

If we approach $(0,0)$ along the line $y=mx$, $$\lim_{(x,y)\rightarrow(0,0)}\frac{x-y}{x+y}=\lim_{x\rightarrow0}\frac{x-mx}{x+mx}=\lim_{x\rightarrow0}\frac{1-m}{1+m}=\frac{1-m}{1+m}$$ Since this result depends on $m$, the original limit does not exist.

I understand that in multivariable calculus, there might be an infinite number of ways to approach (0,0).
But why can't I just plug in the values? And why is it a line, but not a parabola or something else?
I have two more examples,

If we approach $(0,0)$ along the line $y=mx$, $$\lim_{(x,y)\rightarrow(0,0)}\frac{3x^3-y^3}{2x^3+4y^3}=\lim_{x\rightarrow0}\frac{3x^3-m^3x^3}{2x^3+4m^3x^3}=\lim_{x\rightarrow0}\frac{3-m^3}{2+4m^3}=\frac{3-m^3}{2+4m^3}$$ Since this result depends on $m$, the original limit does not exist. Here again we pick y to be a line, but in another example here:

If we approach $(0,0)$ along parabola $y=ax^2$ $$\lim_{(x,y)\rightarrow(0,0)}\frac{x^4+y^2}{x^4-y^2}=\lim_{x\rightarrow0}\frac{x^4+a^2x^4}{x^4-a^2x^4}=\lim_{x\rightarrow0}\frac{1+a^2}{1-a^2}=\frac{1+a^2}{1-a^2}$$ Since this result depends on $a$, the original limit does not exist.

It's a parabola but not a line.
And what if I had some expression under the square root? Would I still pick a line to prove the limit exists or not?
Would just plugging the value (0,0) be wrong in these example?
And would this way (choosing a line or parabola) work if (x,y) were approaching some other points but not (0,0)?

  • $\begingroup$ You can't plug in $x=0$ in $\lim_{x\to 0} \frac{1/x}{\log x}$ either ... $\endgroup$
    – user251257
    Oct 10, 2016 at 23:56
  • $\begingroup$ even in univariate function, you cannot just plug in the number, however, if the function is continuous in the point of consideration, then yes, you can plug the pont in. For the second question, it has not to be a line, you can use parabole or even some random curve in your mind, and yes there are infinitely many curve you can choose. However, you only need to find 2 curves that make the value of the function different, that is why people usually use the simplest curve possible. $\endgroup$ Oct 11, 2016 at 0:36
  • $\begingroup$ For example, in your problem, you can plug $m=1$ when $y=mx$, and the second curve is $y=0$ (the $y$ is already at $0$), and you will find the values are different. Limit does not exist $\endgroup$ Oct 11, 2016 at 0:36

1 Answer 1


You are only allowed to say that $\lim_{x \to a} f(x) = f(a)$ when the function is continuous at $a$ (or for a function of two variables, $\lim_{(x,y) \to (a,b)}f(x,y) = f(a,b)$ ). All of the functions you've given aren't even defined at the point $(0,0)$, and thus cannot be continuous there. So you cannot just plug in $(0,0)$.

As you mentioned, the limit must be the same when we approach from $\textit{every}$ direction. So to show a limit doesn't exist, it suffices to show that a limit is different when we approach from two different directions (just as we find the limits from the left and right in calculus). I can think of two reasons for giving a variety of approaches. For one, it may be the case that we approach along straight lines and get the same limit, but the limit still doesn't exist, and thus need to take a different approach. Also, it just helps to have a few different examples of approaches. As far as what approach you should take, you should start with the easiest approaches - like along the axes - and try different approaches like lines, parabolas, etc. when those easier approaches don't work.

Also, there's nothing special about the point $(0,0)$, and all this works for other points as well.


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