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

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

• You have not squeezed inasmuch as the left-hand side is not bounded below. – Mark Viola Sep 26 '17 at 21:02
• you haven't "sqeezed" your function, unless you can also find a lower bound, and then show that they are equal to one another. – Doug M Sep 26 '17 at 21:02
• And from the diagonal "axis" (where $x=y \equiv z$): $\lim\limits_{z\to0} {|z| \over |z| + |z|} = 1/2$. – David G. Stork Sep 26 '17 at 22:18

## 1 Answer

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.