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$\lim \limits_{x \to 3}$ ${(x^2-2)}$ = 7

So I want to find some $\delta$ > 0 such that for every $\epsilon$ > 0:

$\lvert x^2-9\rvert$ < $\epsilon$ $\iff$ 0 < $\lvert x-3\rvert$ < $\delta$

By a property of absolute value:

$\lvert x+3\rvert$$\lvert x-3\rvert$ < $\epsilon$ $\iff$ 0 < $\lvert x-3\rvert$ < $\delta$

So the left inequality can be written as:

$\lvert x-3\rvert$ < ${\epsilon\over \lvert x+3\rvert}$

Since we're dealing with values of x that are near 3, we can arbitrarily say that we wish to concern ourselves with x in the range:

2 < x < 4

Therefore the maximum value of $\lvert x+3\rvert$ is 7. Since this produces the minimum value of ${\epsilon\over \lvert x+3\rvert}$, we wish to set this as our delta, as it provides the strictest stipulation on the inequality.

$\delta$ = ${\epsilon\over 7}$

Therefore, on the right side of the biconditional statement:

$\lvert x-3\rvert$ < ${\epsilon\over 7}$

We can multiply ${\epsilon\over 7}$ by 7 to get $\epsilon$ on the right side; since the value of $\lvert x+3\rvert$ will never be greater in seven and will also never be negative, the inequality will still hold if we multiply the left side by it.

$\lvert x+3\rvert$$\lvert x-3\rvert$ < $\epsilon$

Simplifying that inequality, we have shown that:

$\lvert x^2-9\rvert$ < $\epsilon$ $\iff$ 0 < $\lvert x-3\rvert$ < $\delta$

I realize that this is most likely frightful. I'm very new to writing proofs of any kind, but it's something I really would like to learn all I can about. Thanks for your time if you choose to respond.

Edit (updated proof):

I want to find some $\delta$ > 0 such that, for some value of $\epsilon$ > 0:

0 < $\lvert x-3\rvert$ < $\delta$ $\Rightarrow$ $\lvert x^2-9\rvert$ < $\epsilon$

Rewriting the right inequality:

$\lvert x-3\rvert$ < ${\epsilon\over \lvert x+3\rvert}$

Restricting x such that:

2 < x < 4

means that the minimum value that ${\epsilon\over \lvert x+3\rvert}$ can attain is ${\epsilon\over 7}$; therefore, the maximum value that $\lvert x-3\rvert$ can obtain is ${\epsilon\over 7}$.

Therefore, setting $\delta$ = ${\epsilon\over 7}$:

$\lvert x-3\rvert$ < ${\epsilon\over 7}$

Multiplying the left side by $\lvert x+3\rvert$ while multiplying the right by 7 is acceptable because $\lvert x+3\rvert$ will always be less than 7, thus preserving the inequality. That leaves us with:

0 < $\lvert x-3\rvert$ < $\delta$ $\Rightarrow$ $\lvert x^2-9\rvert$ < $\epsilon$

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  • $\begingroup$ It is in fact very important to realize that you do not need the "if and only if" at the beginning. It's only that you need to be able to find a delta to give the result for epsilon. Many people think that either it really should be an if-and-only-if, or that it scarcely matters, but the operational point is that in practice it is mostly very hard to find the "exactly right" delta for a given epsilon. But the happy reality is that we don't have to find the optimal/perfect delta for given epsilon. Finding a "too good" delta is completely fine. $\endgroup$ Aug 20 '14 at 23:47
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    $\begingroup$ "So I want to find a $\delta > 0$ such that for every $\epsilon > 0$..." is incorrect. You do not need to find a single $\delta$ which will work for every $\epsilon$. Rather, for a given $\epsilon$ you need to find a $\delta$ (perhaps specific to that $\epsilon$) that works. From the rest of your work I think you recognize that, but it's important to get the wording right. $\endgroup$
    – user169852
    Aug 21 '14 at 0:00
  • $\begingroup$ So if I'm not finding a single $\delta$ which works for every $\epsilon$, how have I actually proved the limit? Wouldn't I need to prove that the inequalities hold for every $\epsilon$ in order to show that as I get infinitely close to the value x is approaching, f(x) gets infinitely close to the limit? $\endgroup$ Aug 21 '14 at 0:51
  • $\begingroup$ @WyattGregory In words, what you need to prove is that you can make $x^2 - 2$ arbitrarily close to $7$ by choosing $x$ sufficiently close to $3$. Putting it another (equivalent) way: you can make $|(x^2 - 2) - 7|$ arbitrarily small by making $|x-3|$ sufficiently small. Translating into mathematics: given any arbitrarily small $\epsilon$, there is some sufficiently small $\delta$ which will ensure that $|(x^2 - 2) - 7| < \epsilon$ provided that $|x - 3| < \delta$. Generally, if I choose a smaller $\epsilon$ I will need a smaller $\delta$ to ensure that that $|(x^2 - 2) - 7| < \epsilon$. $\endgroup$
    – user169852
    Aug 21 '14 at 1:57
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You pretty much have all the right pieces. Note that we're not dealing with a biconditional though; we want to show that: $$ 0 < |x - 3| < \delta \implies |x^2 - 9| < \epsilon $$ Here's a cleaned up version of your proof.


Given any $\epsilon > 0$, consider $\delta = \min\{1, \epsilon/7\} > 0$. Then observe that if $0 < |x - 3| < \delta$, then: \begin{align*} |x^2 - 9| &= |x - 3||x + 3| \\ &< \frac{\epsilon}{7}|x + 3| &\text{since }|x - 3| < \delta \leq \frac{\epsilon}{7} \\ &= \frac{\epsilon}{7}|(x - 3) + (6)| \\ &\leq \frac{\epsilon}{7}\left(|x - 3| + |6|\right) &\text{by the triangle inequality} \\ &< \frac{\epsilon}{7}\left(1 + |6|\right) &\text{since }|x - 3| < \delta \leq 1 \\ &= \epsilon \end{align*} as desired. $~~\blacksquare$

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  • $\begingroup$ As in my comment above... it is very important, to avoid impossibly-difficult enterprises, to not aim for the if-and-only-if, which is completely un-necessary... although it is understandably easy to slip into thinking in those terms in the exaggeratedly formulaic version of "mathematics" many people see ("have to have a formula, and it has to be exactly right...") $\endgroup$ Aug 20 '14 at 23:48
  • $\begingroup$ His proof was basically right,but it's nice to see a cleaned up version by a pro. I'm finding a lot of people teaching basic analysis/honors calculus/advanced calculus (basically all the same course with different audience levels) don't teach explicit epsilon-delta proofs with continuous functions anymore and just use the limit theorems to avoid the work. This is a big mistake to me-students need to be able to do these kinds of "computational" proofs with all kinds of functions.There are going to be examples in the real world where it's not going to be so easy to apply "limit of the sum" etc. $\endgroup$ Aug 21 '14 at 3:43

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