Evaluate limit of as $(x,y)$ approaches $(0,0)$

Question

$$ \lim_{(x,y)\to (0,0)}\frac{|y|}{\sqrt{x^2+y^2}}$$

I know that $|y|$ is both $+y$ and/or $-y$ do I evaluate the limit when $|y|$ is $+y$ and when it is $-y$ seperately to see if the limit matches? If so, what methods can I use?

Answer 1

Consider the trajectory $y=0$, so $$\lim_{(x,y) \to (0,0)} \frac{|y|}{\sqrt{x^2+y^2}}=\lim_{x \to 0} \frac{|0|}{\sqrt{x^2+0}}=0.$$ Now, consider the trajectory $y=x$: $$\lim_{(x,y) \to (0,0)} \frac{|y|}{\sqrt{x^2+y^2}}=\lim_{x \to 0}\frac{|x|}{\sqrt{2x^2}}=\frac{1}{\sqrt{2}}\lim_{x \to 0} \frac{|x|}{\sqrt{x^2}}=\frac{\sqrt{2}}{2}\cdot 1=\frac{1}{2}\sqrt{2}.$$ Since we obtain two distinct values with two distinct trajectories, we conclude that the limit DOES NOT EXIST.

Answer 2

Consider instead using the square root formulation of absolute value for real numbers: $$|y| = \sqrt{y^2}$$

This should give a much more sensible-looking calculation.

Answer 3

Guide:

Consider the path along $x=0$ and $x=y$. What limit do you get along those path?

Answer 4

No, consider the trajectory $x=0$ and the trajectory $y=0$. Then, the limit $L$ that you are working is equal to $1$ because $\sqrt{y^2}=|y|$.

One the other hand, along the path $y=0$ the limit is $L=0$. Hence the limit doesn’t exists, since $0 \neq 1$.