I’ve watched a video on how to prove that $$\lim_{x\to3}x^2=9$$
Using the epsilon-delta definition of a limit, but i have a slightly different proof and I’m wondering whether it is true or false.
The Proof:
Given $\epsilon >0$, let $\delta=\epsilon/6$, suppose $|x-3|<\delta$, then:
$$|x-3|< \frac{\epsilon}{6} \iff 6 |x-3| <\epsilon$$
Since $x$ is very close to $3$ it follows that $|x+3|<6$, which implies that:
$$|x-3| |x+3| <|x-3|6<\epsilon$$
$$|x^2-9|<\epsilon$$
If you think that my argument in $|x+3|<6$ is not rigorous enough you can check it that like this:
$$\delta<x-3<\delta \iff 6-\delta<|x+3|<6+\delta$$
We can make the absolute calues because $6-\delta>0$
Best Answer
Given $\epsilon>0$, we want $\delta>0$ such that
$$|x-3|<\delta\implies |x+3||x-3|<\epsilon$$
$x $ is close to $ 3 $, so we can add the condition $$|x-3|<\color{red}{1}$$ which is equivalent to $$2<x<4 $$ this gives $$5<x+3<7$$
So, $$|x-3|<1\implies$$ $$ |x-3||x+3|<7|x-3|$$ thus we Just need to find $ \delta $ satisfying
$$|x-3|<\color{red}{1}\wedge |x-3|<\delta\implies$$ $$ |x-3|<\frac{\epsilon}{7}$$
From here, we see that we can take $$\delta=\min(\color{red}{1},\frac{\epsilon}{7})$$