Evaluate $\lim\limits_{(x,y)\to(0,0)}\dfrac{|x|}{|x|+|y|}$

The limit does not exist.

From the x-axis (y=0), we have $\lim\limits_{x\to 0} \dfrac{|x|}{|x|+0}=1$

From the y-axis (x=0), we have $\lim\limits_{y\to 0} \dfrac{0}{0+|y|}=0$

Therefore, since $1\neq 0$, the limit does not exist.

But what if we apply the squeeze thereom.

We know that $\dfrac{|x|}{|x|+|y|}\leq \dfrac{|x|}{|x|}=1$, and so the limit is $1$. Why is this incorrect? What are the circumstances that I can use the squeeze thereom, and why isn't this one of them

  • 1
    $\begingroup$ You have not squeezed inasmuch as the left-hand side is not bounded below. $\endgroup$
    – Mark Viola
    Sep 26, 2017 at 21:02
  • $\begingroup$ you haven't "sqeezed" your function, unless you can also find a lower bound, and then show that they are equal to one another. $\endgroup$
    – Doug M
    Sep 26, 2017 at 21:02
  • $\begingroup$ And from the diagonal "axis" (where $x=y \equiv z$): $\lim\limits_{z\to0} {|z| \over |z| + |z|} = 1/2$. $\endgroup$ Sep 26, 2017 at 22:18

1 Answer 1


No, the Squeeze Theorem just gives you in this case that the limit, if it exists, can be at most 1. You really want to use the Squeeze Theorem when you can bound it from both sides and the limit on each side is the same. Otherwise, you really aren't squeezing anything but merely setting 'good boundaries' for your misbehaved limit.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .