Prove by $\epsilon-\delta$ that as $x \to 3,\space 5x\to15$

Question

Prove by an $\epsilon-\delta$ argument that as $x \to 3,\space 5x\to15.$

My workings:

Let $\epsilon>0$ be given such that $\mid \ 5x-15\mid=5\mid x-3\mid<\epsilon$

Thus with $\epsilon=5\delta,\space \mid5x-15\mid<\epsilon$ whenever $\mid \ x-3\mid<\delta$

Therefore the definition of lim$_{x\to3} \ 5x=15\space$ is satisfied. QED

Did I prove this correctly? Thank you guys!

Answer 1

You have a proof sketch but you do not have a proper proof. Follow the steps in this link to write a formal proof.

Suppose $\epsilon > 0$ has been provided. Define $\delta = \frac{\epsilon}{5}$. Since $\epsilon > 0$, then we also have $\delta > 0$.

Now for every $x$, the expression $0 < |x - c| < \delta$ implies $|x - 3| < \frac{\epsilon}{5}$. Can you get to the conclusion from here?