continuity of $f(x)=\frac{xy_0}{x^2+y_0^2}$ at $x_0$

Question

I'm trying to prove $f(x)=\frac{xy_0}{x^2+y_0^2}$($y_0\neq0)$ is continuous at $x_0$ using $\epsilon-\delta$ definition. I tried

$$ \left|\frac{xy_0}{x^2+y_0^2}-\frac{x_0y_0}{x_0^2+y_0^2}\right| = |y_0|\left|\frac{x}{x^2+y_0^2}-\frac{x_0}{x_0^2+y_0^2}\right| =|y_0|\left|\frac{x(x_0^2+y_0^2)-x_0(x^2+y_0^2)}{(x^2+y_0^2)(x_0^2+y_0^2)}\right| \leq|y_0|\left|\frac{x(x_0^2+y_0^2)-x_0(x^2+y_0^2)}{x^2(x_0^2+y_0^2)}\right| =\left|\frac{y_0}{x_0^2+y_0^2}\right|\left|\frac{x(x_0^2+y_0^2)-x_0(x^2+y_0^2)}{x^2}\right| $$

and I don't see how to proceed from here.

Answer 1

at: $|y_0||\frac{x(x_0^2+y_0^2)-x_0(x^2+y_0^2)}{(x^2+y_0^2)(x_0^2+y_0^2)}|$

I would add and subtract the same term, that factors with the terms already there.

$|y_0||\frac{x(x_0^2+y_0^2)-x_0(x_0^2+y_0^2)+x_0(x_0^2+y_0^2)-x_0(x^2+y_0^2)}{(x^2+y_0^2)(x_0^2+y_0^2)}|\\ |y_0||\frac{(x-x_0)(x_0^2+y_0^2)+x_0(x_0^2-x^2)}{(x^2+y_0^2)(x_0^2+y_0^2)}|\\ |x-x_0||\frac{y_0(x_0^2+y_0^2)+x_0y_0(x_0+x)}{(x^2+y_0^2)(x_0^2+y_0^2)}|\\ $

And then show that the factor on the right is bounded.

More abstractly:

if $g(a) \ne 0$ and $f(x),g(x)$ are continuous at $a.$

Then:

$\lim_\limits{x\to a} \frac {f(x)}{g(x)} = \frac {f(a)}{g(a)}$

Answer 2

I'd recommend that instead of $$|y_0|\left|\frac{x(x_0^2+y_0^2)-x_0(x^2+y_0^2)}{(x^2+y_0^2)(x_0^2+y_0^2)}\right|\leq|y_0|\left|\frac{x(x_0^2+y_0^2)-x_0(x^2+y_0^2)}{x^2(x_0^2+y_0^2)}\right|$$ you use $$|y_0|\left|\frac{x(x_0^2+y_0^2)-x_0(x^2+y_0^2)}{(x^2+y_0^2)(x_0^2+y_0^2)}\right|\leq\frac{1}{|y_0|\cdot(x_0^2+y_0^2)}\left|x(x_0^2+y_0^2)-x_0(x^2+y_0^2)\right|$$