show that $f(x,y) = x^2 - y^2$ is continuous at $(0,0)$ using epsilon-delta

Question

show that $f(x,y) = x^2 - y^2$ is continuous at $(0,0)$ using epsilon-delta.

I know that I want $\sqrt{x^2 + y^2} < \delta$ to imply $|x^2 - y^2| < \epsilon$, but I'm not sure how to get there. I tried rewriting things as:

$\left|\frac{(x^2-y^2)(x^2+y^2)}{x^2+y^2} \right| < \epsilon$, and then taking the $1/(x^2+y^2)$ out of the denominator so that it would be

$1/\delta^2\cdot|x^4-y^4| < \epsilon$, but that led nowhere. Any hints/solutions? Thank you!

Answer 1

$| x^2-y^2-0|<\epsilon$ imply $x^2<\epsilon + y^2$ & $y^2<\epsilon + x^2$ Now as $x , y$ are chosen in the nbd of $0$. So $\exists \delta>0$ s.t $|\sqrt {x^2-y^2}|<\delta$[ or if u can show $|x|<\delta$ & $|y|<\delta$ will also enough.] So f is continuous at $f(0,0)=0$.

Answer 2

Let us put $X=(x,y), O=(0,0)$ and $\|X\|=\sqrt{x^2+y^2}$, then: $$|f(X)-f(O)|=|f(x,y)-f(0,0)|=|x^2-y^2| \leq x^2+y^2 =\|X\|^2$$ Let $\varepsilon >0 $ and $\delta=\sqrt{\varepsilon}$, then:$$\forall X \in {\mathbb R}^2 \quad \|X\| < \delta \Rightarrow |f(X)-f(O)| < \varepsilon$$