Prove that $|{xy-x_0y_0}| < \epsilon$

Question

Supposing that $|x-x_0| < min(1, \frac{\epsilon}{2(|y_0|+1)})$ and $|y-y_0| < \frac{\epsilon}{2(|x_0|+1)}$ it follows that $|{xy-x_0y_0}| < \epsilon$

(A friend told me it was helpful to prove this in hopes of proving that $\lim_{x \to c}(f*g)=L*M$ where $\lim_{x \to c}f(x) = L$ and $\lim_{x \to c}g(x) = M$)

I think it's supposed to end something like $|{xy-x_0y_0}| <|x-x_0||y-y_0|<(2(|x_0|+1))*\frac{\epsilon}{2(|x_0|+1)}<\epsilon$ and you're finished, but I get stuck in the steps before.

Answer 1

Hint $$|xy-x_0y_0|=|xy-xy_0+xy_0-x_0y_0| \leq |x| |y-y_0|+|y_0| |x-x_0|$$

Answer 2

just to complete the proof:

$$|x-x_0|<min\left(1, \frac{\epsilon}{2(|y_0|+1)}\right)\leq 1$$ then $|x-x_0| <1$.

$\implies |x|-|x_0| \leq |x-x_0|<1$. Therefore $|x| < 1+|x_0|$.

Also $|y_0|<|y_0|+1$, from the hint we have:

$$|xy - x_0y_0| \leq |x||y-y_0| + |y_0||x-x_0| < (1+|x_0|)\frac{\epsilon}{2(|x_0|+1)} + (|y_0| + 1)\frac{\epsilon}{2(|y_0|+1)} = \epsilon$$