Calculate this limit and find delta the satisfies Calculate: $\lim_{x\to 1}\frac{x^2+2x-3}{x-1}$ and find $\delta \gt 0$ such that:
$|\frac{x^2+2x-3}{x-1} - L | \lt \frac{1}{1000}$
Let's begin:
I first calculated the limit which I found was 4.
let $\epsilon >0 $ 
By the definition of limit we have: 
$$
|\frac{x^2+2x-3}{x-1} - 4 | \lt \epsilon
$$
$$
|\frac{x^2+2x-3-4x+4}{x-1} | \lt \epsilon
$$
$$
|\frac{x^2-2x+1}{x-1}| \lt \epsilon
$$
$$
|\frac{(x-1)^2}{x-1}| \lt \epsilon
$$
$$
|x-1 | \lt \epsilon
$$
This is what I eventually got, but I must say that I haven't understood the structure of the definition of the limit proof, can someone please help me structure the proof and continuing it? 
Thanks.
 A: I agree with you that $4$ is the limit.
Informally, the definition of $x \to a \implies f(x) \to L$ says that for any distance we pick around $L$, we should be able to find a distance around $a$ such that for all $x$ within that distance around $a$ (i.e., for all $x$ *really close to $a$), we will have all of the $f(x)$ within the first distance of $L$ (i.e., we will have all of the $f(x)$ really close to $L$).  In math, this is:
$f(x) \to L$ as $x \to a$ if for every $\epsilon > 0$, $\exists \delta > 0$ such that $|x - a| < \delta$ implies $|f(x) - f(a)|< \epsilon$.  Notice that we are saying for every $\epsilon > 0$, so it should be true as $\epsilon$ gets smaller and smaller, that we will always be able to find some distance around $a$ such that all of the elements $x$ within that distance will have $f(x)$ within $\epsilon$ of $L$.
Now, the work you did was kind of working "backwards" to find the $\delta$, and this is very common in $\epsilon$-$\delta$ proofs.  You said: suppose we have that $|\dfrac{x^{2} + 2x - 3}{x - 1} - 4| < \epsilon$, i.e., the distance between $f(x)$ and $4$ is less than $\epsilon$.  Well, if that's true, then your next line should be true, and if that is true then the next line after that should be true, etc.  You got all the way down to $|x - 1| < \epsilon$.  So that means if it is true that $|x - 1|< \epsilon$ (i.e., the distance between $x$ and $1$ is less than $\epsilon$), then it's true that our function is less than $\epsilon$ away from $4$.  So in your proof, the $\delta$ you found was actually $\delta = \epsilon$.
