# Could I get a critique of this epsilon-delta limit proof?

$\lim \limits_{x \to 3}$ ${(x^2-2)}$ = 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 < $\lvert x-3\rvert$ < $\delta$ $\Rightarrow$ $\lvert x^2-9\rvert$ < $\epsilon$

• 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. Aug 20 '14 at 23:47
• "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.
– user169852
Aug 21 '14 at 0:00
• 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? Aug 21 '14 at 0:51
• @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 $|(x^2 - 2) - 7| < \epsilon$.
– user169852
Aug 21 '14 at 1:57

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. $~~\blacksquare$