[Math] $\epsilon$-$\delta$ proof of a limit of a function $f(x,y)$

Question

I was working on this exercise to prove the differentiability of a function at a certain point, but I got stuck in proving the following limit.

$$\lim_{(x,y)\rightarrow (0,0)} \frac{2x^2-3y^2}{\sqrt{x^2+y^2}} = 0 \>.$$

I'm still getting used to deriving the delta-epsilon proof of a limit for functions of many variables, any help is really appreciated!

Answer 1

You have $$ |f(x,y)|\leq \frac{2x^2+3y^2}{\sqrt{x^2+y^2}}\leq \frac{3x^2+3y^2}{\sqrt{x^2+y^2}}=3\sqrt{x^2+y^2}. $$ Now you can apply the $\epsilon-\delta$ procedure.