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Claim: $$\lim_{x\to 1} \frac{100}{x} = 100$$

Proof: Let any $\epsilon > 0.\ |\frac{100}{x}-100| = 100 |\frac{1}{x}-1| = 100 \frac{|x-1|}{|x|}.$

If this term is smaller than epsilon in the interval where $0<|x-1|<\delta$, then we are done.

Assume $\delta=100.$ Then, $|x-1|<100, -99<x<101, -1/99<1/x<1/100$

Thus $100 \frac{|x-1|}{|x|} <\frac{ 100}{100*|x-1|}$

So if $0<|x-1|< \min\{\epsilon, 100\}=\delta,$ then $|\frac{100}{x}-100| < \epsilon$. QED

Note: Even if this is correct, my intent of asking this question was because I'm having doubts on the fundamental logic lying behind this proof... For some reason, I don't like the fact that we can let delta be something and then later "rectify" it after putting in a form of epsilon... Like I can let delta be less than 1/2 and get (I think) for delta = min{epsilon/200, 1/2} and that's different from what I got... I know that epsilon is any number, but it just seems weird. Did I misunderstand something?

I will be really grateful if someone can help me in any way understand this better.

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Please check the way I interpreted your equations, especially the line starting Thus. I wasn't sure if the $|x-1|$ should be in the denominator. – Ross Millikan Oct 19 '11 at 14:12
Often the choice of $\delta$ is decided in private and simply chosen without fanfare in the proof itself. – Hans Parshall Oct 19 '11 at 14:17
@HansParshall I think I have understood what you meant. Can you read the comment I made on Willie Wong's post below to check! Much thanks! – MathMathCookie Oct 19 '11 at 21:27
It looks like you get it to me. I'm glad this helped. – Hans Parshall Oct 19 '11 at 22:48
up vote 5 down vote accepted

You made a mistake in the implication that

$$ -99 < x < 101 \implies -\frac{1}{99} < \frac{1}{x} < \frac{1}{100} $$

When $A, B$ are positive numbers, you have (note the reversal of inequality signs)

$$ A < x < B \implies \frac{1}{A} > \frac{1}{x} > \frac{1}{B}$$

When $A,B$ are both negative numbers, you also have (the same thing)

$$ A < x < B \implies \frac{1}{A} > \frac{1}{x} > \frac{1}{B} $$

But when $A < 0 < B$, all you can say is

$$ A < x < B \implies \frac{1}{A} > \frac{1}{x} \mbox{ or } \frac{1}{B} < \frac{1}{x} $$

in particular, the absolute value

$$ \frac{1}{|x|} > \min( \frac{1}{|A|}, \frac{1}{|B|} )$$

is not bounded above a priori.

To address your query

For some reason, I don't like the fact that we can let delta be something and then later "rectify" it after putting in a form of epsilon

perhaps it would be clearer if the $\epsilon$-$\delta$ statements were written this way (I'll use the case of the limit as an example, but you can substitute similar statements into other definitions):

Given a function $f(x)$ and a point $x_0$, and a value $y$. We say that $\lim_{x\to x_0} f(x) = y$ if there "we can find" a function $\delta(\epsilon)$ (a function $\delta$ that depends on the variable $\epsilon$) such that the expression $|f(x) - y| < \epsilon$ holds true whenever $0 < |x-x_0| < \delta(\epsilon)$.

Rmk: I put "we can find" in quotes because in non-school mathematics (as practised by professional mathematicians), we usually cannot write down an explicit formula for the function $\delta(\epsilon)$. We just can demonstrate that such a function must exist.

In particular, doing an $\epsilon$-$\delta$ proof is like reducing a system of algebraic equations: you want to "solve" for the function $\delta(\epsilon)$ (subject to the above caveat).

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I believe I have found my fault then! By setting delta to be such a large value, I have made A<0, and thus 1/|x| is not bounded above. Which is useless to me since I need to use the fact that 1/|x| is bounded above to continue on with my proof! So, I think I finally understood why delta can not be some absurd value. You must choose delta using trial and error until you find something nice enough so that we get some sort of 1/A>1/x and then use that in your proof. Right? And now I think I understand what Hans Parshall meant above as well. – MathMathCookie Oct 19 '11 at 21:25
@MathMathCookie : almost! All you wrote is correct philosophically. But in practice we don't "choose $\delta$ using trial and error", our choices are guided by algebraic principles and/or past experience, the same way that when I ask you to solve the equation $4x^2 + 15 x - 3 = 0$, you generally don't proceed by trial and error, but by either applying the quadratic formula (something from "experience") or by manually completing the square and taking the square root (something from "principles"). – Willie Wong Oct 19 '11 at 23:04
This is how I would "reason" in your case. Start with the inequality we want: $|x-1|/|x| < \epsilon / 100$ (which is rearranging what you wrote on the first line of your proof). Next we say to ourselves: if $x$ is very close to one, then $|x-1| / |x|$ is almost $|x-1|$. The solution to $|x-1| < \epsilon / 100$ is $1 - \epsilon/100 < x < 1 + \epsilon/100$. But if $x < 1$, the approximation of $|x-1| / |x|$ by $|x-1|$ is a bad one. So we cannot actually go all the way down to $1 - \epsilon/100$. But if $x$ is almost $1$, it is certainly at least $1/2$. So we replace our approximation – Willie Wong Oct 19 '11 at 23:12
$|x-1|/|x| \approx |x-1|$ by an actual inequality $|x-1|/|x| < 2|x-1|$ provided $|x| > 1/2$, which can be guaranteed if $|x-1| < 1/2$. So in the end we want simultaneously to satisfy $|x-1| < 1/2$ and $2|x-1| < \epsilon/100$. This can be achieved if $|x-1| < \min (1/2, \epsilon/200)$. --- For the above I wrote precisely as I think, so it looks a bit messy. In writing the proof part of the work is to take the "stream-of-conciousness" and convert it to a more streamlined description of your argument. – Willie Wong Oct 19 '11 at 23:17
One last little bit of resource is the Tricki. The above train of thought corresponds with the advice "If an argument looks promising but needs some technical hypothesis, try assuming that hypothesis for now, but aim to remove it later", together with using Technique 3 from this page. – Willie Wong Oct 19 '11 at 23:25

You (correctly) want to make $100 |x - 1| / |x| < \epsilon$ and you chose the (unfortunately, incorrect) strategy of making $|x - 1| < \epsilon$ and $100 / |x| < 1$ for $x$ near $1$. The reason this won't work is that when $x$ is close to $1$, you can't expect $100 / |x| < 1$!

So instead let's try to make $|x - 1| < \epsilon / 200$ and $100 / |x| < 200$. Now, we can definitely satisfy the second equality, which is equivalent to $|x| > 1/2$, when $x$ is near $1$. In fact, if $|x - 1| < 1/2$, then $|x| > 1/2$, so this tells us we should choose $\delta < 1/2$. (Where did I pick $200$ from? Well, I just wanted a number big enough to get this second inequality to work out. We'll see below that putting the $200$ in the denominator of the first inequality won't cause us a problem.)

To satisfy the first inequality, all we need to do is choose $\delta < \epsilon / 200$. Then certainly, if $|x - 1| < \delta$, we have $|x - 1| < \epsilon / 200$.

Since we need both $\delta < 1/2$ and $\delta < \epsilon / 200$, we need to choose $\delta < \min(1/2, \epsilon/200)$.

BTW, don't be discouraged that you went down a wrong path in trying to solve your problem. The way to become better at math is to try hard problems and after you inevitably stumble down a wrong path, to go back and figure out why that path didn't work.

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Your proof is inaccurate: if $|x-1|<100$ you conclude that $\frac{1}{x}$ is in $(\frac{-1}{99}, \frac{1}{100})$. That's not true, for example, we could take $x$ to be very small, say $\frac{1}{1000}$. Clearly the first inequality holds, however $\frac{1}{x}$ is very large, and not in the small domain above. Here instead of taking $\delta$ to be something large like $100$, we need to avoid this singularity of $\frac{1}{x}$, so we should make sure $x$ doesn't vanish. Let's take $\delta = \frac{1}{2}$, that way if $|x-1| < \delta$ then we indeed bound $\frac{1}{x}$ to be in the domain $(\frac{2}{3}, 2)$.

We now have $\frac{100|x-1|}{|x|} < 100 * 1/2 * 2$, so if we take a revised $\delta$ of $min({1/2, \epsilon/200})$ we conclude $100|x-1|/|x| < 200*\epsilon/200 = \epsilon$.

As per the confusion of how the value of $\delta$ changes, it's not that the value of $\delta$ changes, so much as we take some $\delta$ and then later on set its value. So initially we can take a generic $\delta$ that we haven't set yet, and as a variable, see how the expression we're analysing changes.

Consider the first part of the proof to be "if $\delta$ were this value then..." and then we discover that we need a slight refinement, so we consider $\delta$ to still hold the constraint we made in the first part, but add a new constraint, in order to meet another condition.

So we found that if $\delta = \frac{1}{2}$ then $100-\frac{100}{x}$ is bounded by a constant multiple of $\delta$, and as a result if we make sure that $\delta$ will still be at most a half, but also no more than $\epsilon$, say we choose $\delta = min({\frac{1}{2}, \epsilon})$ then we prove the statement. So essentially we only choose once, at the end. Beforehand we merely watched what happens if certain conditions on $\delta$ were met.

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Let us first correct some slight mistakes in your proof

1- Line Assume : you get $1/101<1/x<1/99$

2- Line Thus : you get $100\frac{|x-1|}{|x|}<\frac{10000}{99}$

So you do not get very far (meaning very close to $0$) with this choice of $\delta$.

The definition of continuity is to be sure that by making $x$ is close to $1$ you can go as close as you want to $0$ for the quotient you are analyzing.

So you begin by fixing $\epsilon$ small (like $<1$). Then you try to find a $\delta$ that make the quotient close to $0$ less than $\epsilon$. You can find it by trial and error, but usually the $\delta$ will depend on $\epsilon$.

Here if you choose for instance $\delta<\frac{\epsilon}{200}$ I think it will work right?

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