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I need to prove that the limit of $x^2$ as $x$ approaches 3 equals 9.

For a given $ > 0$, there exists a $ > 0$ such that:

$$0 < |x - 3| < ⇒ |x² - 9| < $$

Taking the epsilon inequality I started the scratchwork.

$$|x² - 9| = |(x + 3)(x - 3)| < |x + 3| = $$

so we have $ = /|x + 3|$

I know that I have to bound the $|x + 3|$ to find a constant and write $$ in terms of $$, so I let $ = 1$ and:

$$|x - 3| < 1$$

$$\Rightarrow -1 < x - 3 < 1$$

$$\Rightarrow 5 < x + 3 < 7$$

$$\Rightarrow |x + 3| < 7$$

$$ \Rightarrow = /7$$

It seems to be right,but it was told me that I'm not setting $ = 1$, but $ < 1$. Why? My point here is not "why 1 and not any other number" but why the inequality sign and not the equal sign? Letting $ = 1$ wouldn't still be useful because $|x - 3|$ is still less than 1?

ZYX
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LightL96
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  • It does not matter much. In fact the $\epsilon-\delta$ argument would also work with $0 < |x - 3| \le \delta⇒ |x^2 - 9| \le \epsilon$. But you probably want to use $0 < |x-c|$ and avoid $0 \le |x-c|$ since $\lim\limits_{x \to c} f(x)$ does not care about $f(c)$, and so it may be easier to use $<$ through out – Henry Dec 08 '23 at 00:44
  • Wonderful. Thank you for the explanation. So using the δ < 1 is just a matter of detail when dealing with |x - 3|? – LightL96 Dec 08 '23 at 00:50
  • The $1$ matters here because you want to justify the $7$. The $<1$ rather than $\le 1$ is a detail to reduce the risk of writing $0 \le |x-3|$. – Henry Dec 08 '23 at 00:57
  • How would this risk show up to me in an example? – LightL96 Dec 08 '23 at 01:02
  • What is the limit of $\dfrac{x^3-3x^2}{x-3}$ as $x \to 3$? If you were to consider $0 \le |x -3|$ then you would have to consider the possibility of $0=|x-3|$ i.e. $x=3$, but $\dfrac{x^3-3x^2}{x-3}$ is not defined when $x=3$ as it would be $\dfrac00$. So you might want to avoid any $\le$ signs. – Henry Dec 08 '23 at 01:08
  • Now I see. It was just confusing because I still watch some videos on YouTube, sometimes they write "let δ = 1 and then |x - 3| < 1", some say δ < 1 without giving any explanation. Thank you! – LightL96 Dec 08 '23 at 01:26
  • In fact, if you first fix $0<\delta\leq 2023$, and then later require $\delta<\frac{\epsilon}{2023+2(3)}=\frac{\epsilon}{2029}$, i.e if you fix any number $\delta$ such that $0<\delta<\min\left(2023,\frac{\epsilon}{2029}\right)$, things would still work out. (My point isn’t just with the 2023, it is with the inequalities). See here for a little bit more on the thought process. Also, this ties in to something you will undoubtedly realize as you gain familiarity: analysis is more about inequalities than equalities :) – peek-a-boo Dec 08 '23 at 01:42
  • Thank you so much! It will help me a lot. – LightL96 Dec 08 '23 at 01:55

1 Answers1

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I think you're misinterpreting what was said to you. You can't be setting $\delta = 1$ because you're ending up with $\delta = \epsilon/7$, and that would contradict your setting of $\delta=1$. So then what do you even mean when you say that you have set $\delta$?

What that chain of calculations is saying is that "When $\delta=1$ does not work, then $|x-3|<1$, and so then $\delta=\epsilon/7$ should work." That's why the answer is the minimum of $1$ and $\epsilon/7$. The value of $\delta=1$ would not work when $\epsilon=0.01$ and the value of $\delta=\epsilon/7$ would not work when $\epsilon=1000$.

SuhailSherif
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