I am having some issues with the following multivariable limit:

$$\lim_{x,y\to0,0} \frac{x^2+y^2}{x+y}$$

I am trying to show whether it exists and is equal to 0, or whether it does not exist. What I tried to do was convert it to polar coordinates and then show the limit was zero from there, however I am not sure if what I have done is valid. Here is how I have done it:

Rewrite in terms of polar coords: $$\lim_{r\to0^+} \frac{r^2}{r\sin(\theta) + r\cos(\theta)} = \frac{r}{\sin(\theta) + \cos(\theta)}$$

We know that for all $\theta$ that does not make the denominator $0$ (i.e everywhere in the domain of the function), this limit must be $0$. Is this correct? More specifically, is $\lim_{r\to0^+}$ replacable with $\lim_{x,y\to 0,0}$ ?

  • $\begingroup$ Switching to polar coordinates is one trick, but not the only trick that helps with multivariate limits. In this case, when $y\to -x$, the function is arbitrarily large. Hence the limit does not exist. $\endgroup$
    – vadim123
    Nov 30, 2015 at 0:59
  • $\begingroup$ @vadim123 Ok, but $y = -x$ is not in the domain of the function so we cannot approach it along that path. Right? $\endgroup$ Nov 30, 2015 at 1:18
  • $\begingroup$ How does $y = -x$ help? Wouldn't that make the divisor 0? $\endgroup$ Sep 19, 2017 at 15:21

2 Answers 2


If we approach the origin along the line $y=kx$ the limit is $0$ : $$\lim_{(x,y)→(0,0)}\frac{(x^2+y^2)}{(x+y)}=\lim_{(x,kx)→(0,0)}\frac{(x^2+(kx)^2)}{(x+kx)}=\lim_{(x→0)}\frac{(x(1+k^2))}{(1+k)}=0$$ If we approach the origin along the curve $y=ax^2+bx$ we have : $$\lim_{(x,y)→(0,0)}\frac{(x^2+y^2)}{(x+y)}=\lim_{(x,ax^2+bx)→(0,0)}\frac{(x^2+(ax^2+bx)^2)}{(x+ax^2+bx)}= \lim_{(x→0)}\frac{x(1+b^2+2abx+a^2 x^2)}{(1+b+ax)}\to(1)$$ Now , if $b=+1$ the limit is again $0$. If $b=-1$ expression (1) becomes : $$\lim_{(x→0)}\frac{x(2-2ax+a^2 x^2)}{ax}=\lim_{(x→0)}\frac{2-2ax+a^2 x^2}{a}=\frac2a$$ So if for $b=-1$ we let, say $a=1$, we can choose to approach the origin along the curve $y= x^2-x$ and we get $$\lim_{(x→0)}\frac{2-2ax+a^2 x^2}{a}=\lim_{(x→0)}{(2-2x+x^2)}=2$$Since two different limit values on different path approach, the limit does not exist.

  • 3
    $\begingroup$ Please use MathJax to format your answer. It is very hard to read (although it does look correct). $\endgroup$
    – Joey Zou
    Sep 28, 2016 at 22:25

Let a curve $C$ be described parametrically as $x=t$ and $y=-t+t^3$. Note that the curve approaches the origin as $t\to 0$.

We have on $C$


Thus, on $C$ the limit of interest is undefined since the limit from the right is

$$\lim_{t\to 0^+}\left(t^3-2t+\frac2t\right)=\infty$$

while the limit from the left is

$$\lim_{t\to 0^-}\left(t^3-2t+\frac2t\right)=-\infty$$

We conclude that the limit

$$\lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{x+y}\,\,\text{does not exist}$$

  • $\begingroup$ Thanks, I was wondering, how do you know to use/justify such a parametrization? Is it just arbitrary? $\endgroup$ Nov 30, 2015 at 13:53
  • 1
    $\begingroup$ Great question. Observe that the denominator was $x+y$, so one is compelled to try something where $y=-x+f(x)$, where $f$ approaches $0$ more rapidly than the terms in the numerator. $\endgroup$
    – Mark Viola
    Nov 30, 2015 at 15:26
  • $\begingroup$ Dr. MV - can you recommend any texts that offer a good exposition of multivariate limits? I have only found brief treatments that make their evaluation seem like a second cousin to divination... $\endgroup$
    – Rax Adaam
    Nov 7, 2016 at 22:13
  • $\begingroup$ Rax. I am unaware of any text that devotes an extensive amount of time to that subject. $\endgroup$
    – Mark Viola
    Nov 7, 2016 at 23:28

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