Epsilon delta for absolute value proof

Question

If $x,z$ $\epsilon$ $R$, show that for every $\epsilon >0$ there is a $\delta > 0$ such that if $y$ $\epsilon$ $R$ satisfies $|y-x|< \delta$ then $|zy - xz| < \epsilon$.

So I tried to take $\epsilon =1$ and I tried to solve from there, but I couldn't seem to make it work. Any ideas/help?

Answer 1

if $z = 0$, the statement is a triviality. Hence, we assume $z \neq 0 $.

Let $\epsilon >0 $ be given. Take $\delta = \frac{\epsilon}{|z|} $. Let $y \in \mathbb{R}$ be arbitrary and let us suppose that $|y -x | < \delta$. It follows that

$$ |zy - zx| = |z(y-x)| = |z||y-x| < |z| \cdot \frac{ \epsilon}{ |z|} = \epsilon$$