proving that $\forall(\epsilon>0)\exists(\delta>0)\forall(x\geq1)\forall(y\geq1)(|x-y|<\delta \implies|\frac {1}{x}-\frac{1}{y}|<\epsilon)$

Question

I'm having some hard time with formal proofs, and was wandering if my solution is decent (I mostly write my proofs in Hebrew, so please let me know if there something off with the English here).

Let $\epsilon>0, x\geq1, y\geq1.$

Suppose there's $\delta>0$ so $|x-y|<\delta$

Apply power -1 to both sides of the inequation. Now we have: $$|x^{-1}-y^{-1}|<\delta^{-1}=|\frac {1}{x}-\frac{1}{y}|< \frac{1} {\delta}$$

Choose $\delta = \epsilon^{-1}$ and we'll have:

$|x-y|<\delta=\epsilon^{-1} \implies|\frac {1}{x}-\frac{1}{y}|<\epsilon$

$\square$

Answer 1

It's wrong. First, it makes no sense to supposer that there's such a $\delta$; you are supposed to find one. And it is false that$$|x-y|<\delta\implies\left|\frac1x-\frac1y\right|<\frac1\delta.$$

Answer 2

Note that $|a-b|^{-1} \ne |a^{-1}-b^{-1}|$, hence the working is not correct.

Note that we have

$$\left|\frac1x-\frac1y \right|=\frac{|x-y|}{|xy|} \le |x-y|$$

since $x,y\ge1$.

Hopefully you can see how to choose the $\delta$.