# Question regarding epsilon-delta proofs.

1. Oct 12, 2013

### TheFerruccio

1. The problem statement, all variables and given/known data

Prove that if $\lim_{x\to a}f(x)=A\neq 0$ then $\lim_{x\to a}1/f(x)=1/A$
2. Relevant equations
This is a proof, so it's just an epsilon delta proof. I know the solution. I am asking about a thought process.
3. The attempt at a solution

Pick a $\delta_1$ small enough such that if $\left|x-a\right|<\delta_1$ then $f(x)>A/2$

Then...

$\left|\frac{1}{f(x)}-\frac{1}{A}\right|=\left|\frac{f(x)-A}{f(x)A}\right|<\frac{2}{A^2}\left|f(x)-A\right|$

Now that it is in the form of $\left|f(x)-A\right|$ Now choose an $\epsilon=\frac{A^2}{2}\epsilon$ such that if $\left|x-a\right|<\delta$ then $\left|f(x)-A\right|<\frac{A^2}{2}\epsilon$ then $\left|\frac{1}{f(x)}-\frac{1}{A}\right|=\frac{2}{A^2}\frac{A^2}{2} \epsilon = \epsilon$ This completes the proof.

I understand why this proof works, but I do not understand the thought process to getting to this proof. There is some sort of creativity or accuracy in guessing an appropriate value for the choice of epsilon that completely eludes me. I've struggled over this for weeks and I have not had much leeway at all. If you were to begin this proof, how would you go about approaching the problem to choose the appropriate values for epsilon, and the first limit? I am specifically focusing on why A/2 was chosen (but I understand why A/3 or 3A/4 would also work, and why that would alter the subsequent epsilon). If I were just starting this problem out, what procedure would I pursue in my head to organize this and come up with this solution as a proof?

2. Oct 13, 2013

### brmath

Here is something more straightforward and once you've looked at it you will have better insight into the A/2 etc.:

Since $\lim_{x \rightarrow a} f(x) = A$ then for any $\epsilon$ > 0 there exists $\delta$ > 0 such that when |x - a | < $\delta$ then |f(x) - A| < $\epsilon$. That last inequality is the same as:

A -$\epsilon$ < f(x) < A + $\epsilon$. So $\frac {1}{A - \epsilon} > 1/f(x) > \frac {1}{A + \epsilon}$. Then $\frac {1}{A - \epsilon} -1/A > 1/f(x) - 1/A > \frac {1}{A + \epsilon} -A$

Adding things up we get $\frac {\epsilon}{A (A- \epsilon)} > 1/f(x) - 1/A > \frac {-\epsilon}{A(A + \epsilon)}$

Then |1/f(x) - 1/A| < $k \epsilon/A^2$ where k>0 is some constant to deal with the fact that I omitted the $\epsilon$ in the denominator.

3. Oct 13, 2013

### TheFerruccio

I understand everything you did until the last step. Why did you choose to eliminate the epsilon in the denominator? I also still don't see the connection to the A/2 at all. Thanks for the help.

4. Oct 13, 2013

### brmath

A + $\epsilon$ is very close to A if $\epsilon$ is small. Rather than letting it linger around in the denominator, I called the denominator $A^2$. Now what did that do to the inequality?

$(A-\epsilon)(A+\epsilon) = A^2 - \epsilon^2$ which is a little smaller than $A^2$. So the entire fraction will be larger than if I were dividing by $A^2$. To reduce the fraction down to the same level, I threw in a k < 1 to knock the fraction back down to what it would have been had I not eliminated the $\epsilon 's$; k does depend on $\epsilon$, but it always exists.

Now the A/2. He wrote it rather sloppily -- he should have said |f -A| >| A/2|. What he means is to start where f is beginning to close in on A. The 2 is kind of arbitrary, and probably serves the same purpose as my k. That is, it allows us to be a bit sloppy with the inequalities so as to make the final answer neater.

I now remember that using A/2 is a standard way of doing this problem, and there are probably lots of books which do it that way.. It makes things a little neater than the way I did it. But as you say, it is not intuitive.

5. Oct 13, 2013

### LCKurtz

To add to other's comments, since your question is about the thought process, I will add my 2 cents worth. I always start such problems with an "exploratory" argument. We want to make$$\left| \frac 1 {f(x)} - \frac 1 A\right|$$ small. Rewriting it as$$\left| \frac{A - f(x)}{Af(x)}\right|=\frac{|A-f(x)|}{|A||f(x)|}$$we know we can make the numerator small and hence the whole fraction small if we can keep the denominator away from $0$. Luckily we are given $A\ne 0$. So the problem becomes keeping $f(x)$ away from $0$. We can do this by picking $\delta$ small enough so that $f(x)$ is within $\frac {|A|} 2$ of $A$, hence away from $0$. Now the problem becomes additionally picking $\delta$ small enough to make it all come out less than $\epsilon$. That's the easy part once you get this far. When you are done you write the thought process out backwards, making the reader wonder "why did he pick that $\delta$?".

6. Oct 14, 2013

### TheFerruccio

Ah, so it is a case of working backwards! That last reply helps a ton with my understanding of it. So, I make a statement along the lines of "this is what I want to end up with" then I explore the base functions (for instance, if the form has f(x)g(x) in it, I'd want to get it in the form of f(x)-a and g(x)-b. Sorry for my poor wording, but I think I am beginning to understand the process. It is definitely not an intuition that is easily seen once compacted into simple solutions that fit in books.

7. Oct 14, 2013

### brmath

Working backwards is a time honored approach. And as you point out, then when you work it forward, you can clean and tune up, so that eventually the original logic becomes obscured.