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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 $6-\delta>0$

PNT
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    "because $6-\delta>0$" Not necessarily. What if $\epsilon=37$? – Arthur Apr 03 '21 at 19:11
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    "Since $x$ is very close to $3$ it follows that $|x+3|<6$," what about $x = 3.00001?$ – user2661923 Apr 03 '21 at 19:42
  • Personally, I start with $|x - 3| < \delta \implies -\delta < (x-3) < \delta \implies (3-\delta) < x < (3+\delta).$ Then, I will typically add an artificial constraint (e.g. $\delta \leq [1/2]$), and then derive a separate constraint on $\epsilon$ (e.g. $\delta < [\epsilon/7]$) or something like that. Then, I will put it all together via $\delta \leq \min([1/2], [\epsilon/10])$, or something that looks like that. This will be my candidate specification. Then, I will manually verify that my candidate specification works against the definition of a limit converging to a value. – user2661923 Apr 03 '21 at 19:49
  • The artificial constraint $(e.g. \delta \leq [1/2]$) is nothing more than a contrivance that allows me to assume that $\delta^2 < \delta.$ This facilitates my specifying a linear relationship between $\delta$ and $\epsilon.$ – user2661923 Apr 03 '21 at 19:51

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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 $$\delta=\min(\color{red}{1},\frac{\epsilon}{7})$$

hamam_Abdallah
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