is my epsilon-delta proof correct?

Question

I am a complete beginner with limits and I self study, so I don't have anyone to confirm my answers. I had a simple limit to prove with the precise definition, its a linear equation and I did lots of these already but I want to be sure because I don't feel like I am ready to continue. Here is what I did:

$$ \lim_{x \to -3}(7x-9)=-30 $$ $$ |7x-9+30|< \epsilon $$ $$ 7|x+3|< \epsilon $$ $$ |x+3| < \frac{\epsilon}{7} $$ $$ \bbox[1px, border: 1px solid black] { \delta = \frac{\epsilon}{7} } $$

---

and with the delta I can move on to the formal proof:

$$ 0<|x+3|<\delta=\frac{\epsilon}{7} $$ $$ 0<7|x+3|<7\cdot \frac{\epsilon}{7} $$ $$ 0<|7x+21|<\epsilon $$ $$ 0<|7x-9+30|<\epsilon \ \ \blacksquare $$

---

Is this correct?

Answer 1

For a rigorous proof :

We have that $$|(7x-9)+30|=7|x-3|.$$

Let $\varepsilon >0$. Set $\delta =\frac{\varepsilon }{7}$. If $|x-3|\leq \delta $, then $$|(7x-9)+30|=7|x-3|\leq 7\cdot \delta =7\cdot \frac{\varepsilon }{7}=\varepsilon .$$ Since $\varepsilon >0$ is unspecified, the claim follow.

Answer 2

It's good. I particularly like the way you distinguish between the proof and the calculation. On that point, when proving, you can go a step further and omit the calculation. Simply start by assuming $\varepsilon > 0$ and $0 < |x + 3| < \frac{\varepsilon}{7}$. Then, your steps show that $|7x - 9 + 30| < \varepsilon$, proving $\lim_{x \to -3} (7x - 9) = -30$.