What's wrong with my epsilon delta proof here? 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.
 A: 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).   
A: 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.
A: 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.
A: 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?
