I am trying to prove that a function $\ f(x,y)=y^2x$ is continuous everywhere, using epsilon delta proofs.
I've set it up like:
$\ |x^2y-a^2b| < \epsilon $
and
$\ \sqrt{\left(x-a\right)^2+\left(y-b\right)^2} < \delta $
But other than that, have absolutely no clue about what operations I need to do to the top equation. If someone could please point me in the right direction that would be great.
Fix $(a,b)\in \Bbb R^2$. For all $(x,y)\in \Bbb R^2$,
\begin{align}|f(x,y) - f(a,b)| &= |y^2x - b^2 a|\\ & = |(y^2x - y^2 a) + (y^2a - b^2 a)|\\ & \le |y^2x - y^2a| + |y^2a - b^2a|\\ & = y^2|x - a| + |y^2 - b^2||a|. \end{align}
Now if $(x - a)^2 + (y - b)^2 < 1$, then in particular $|y - b| < 1$. So by the triangle inequality, $|y| \le |y - b| + |b| < 1 + |b|$ and $|y + b| \le |y - b| + 2|b| < 1 + 2|b|$. Thus
\begin{align}y^2|x - a| + |y^2 - b^2||a| &= y^2|x - a| + |a||y + b||y - b| \\ &<(1 + |b|)^2|x - a| + (1 + 2|b|)|a||y - b|. \end{align}
The last expression can be made less than a positive number $\epsilon$ by having $|x - a|$ and $|y - b|$ less than
$$\frac{\epsilon}{(1 + |b|)^2 + |a|(1 + 2|b|)},$$
which can be done by having
$$\sqrt{(x - a)^2 + (y - b)^2} < \frac{\epsilon}{(1 + |b|)^2 + |a|(1 + 2|b|)}.$$
Hence, given $\epsilon > 0$, set
$$\delta = \min\left\{1, \frac{\epsilon}{(1 + |b|)^2 + |a|(1 + 2|b|)} \right\}.$$
For all $(x,y)\in \Bbb R^2$, $(x - a)^2 + (y - b)^2 < \delta^2$ implies
$$|f(x,y) - f(a,b)| < (1 + |b|)^2|x - a| + (1 + 2|b|)|a||y - b| < \epsilon.$$