I'd like to share a worked example of an epsilon - delta proof of limit found at: http://www.karlscalculus.org/x2_1.html and reproduced here for your convenience. Karl Hahn explains each step of the proof so even a beginner at proofs like can fully understand it.
Worked Example
Prove the following limit:
$$\lim_{x → 2}\frac{2 (x^2 - 4)}{(x - 2)} = 8 \qquad\qquad\text{eq. 2.1x-1}$$
using the delta-epsilon method.
Clearly we cannot evaluate this function at $x = 2$ because that would make for a zero denominator.
We recall from algebra that $x^2 - 4$ is the difference of squares and can therefore be readily factored.
$$x^2 - 4 = (x + 2) (x - 2)$$
Let's give the function we are trying to find the limit of a name. Let's call it $f(x)$. So we have:
$$f(x) = \frac{2 (x^2 - 4)}{(x - 2)} = \frac{2 (x + 2) (x - 2)}{(x - 2)} \qquad\qquad\text{eq. 2.1x-2}$$
Clearly at all values of $x$ except $x = 2$ we get a cancellation, and this is the same as:
$$f(x) = 2 (x + 2) \qquad\qquad\text{eq. 2.1x-3}$$
So the above holds for any value of $x$ except $2$. That means we can evaluate $f(x)$ using this expression and get the right answer provided x is never equal to 2. So if we prove that:
$$\lim_{x → 2} \;\;2 (x + 2) = 8 \qquad\qquad\text{ eq. 2.1x-4}$$
then we have solved the problem. Do you understand why? Remember that all the expressions we have made for $f(x)$ are identical functions for any value of $x$ besides $2$. Taking the limit of $f(x)$ as $x$ goes toward $2$ means never having to evaluate $f(x)$ at $x = 2$. So at all the places we do have to evaluate it, all the forms of $f(x)$ are indeed identical, including $f(x) = 2 (x + 2)$.
Now we get down to the delta-epsilon part of the proof. We have to arrange a scheme by which you can tell me how close $f(x)$ has to be to $8$ (i.e. you give me an $ε$) and based upon that can tell you how close $x$ has to be to $2$ to make it true (i.e. I can give you a $δ$ that makes it true).
Well, let's just set up an equation that shows what happens when we use an x that is within δ of 2.
$$f(2 ± δ) = 2 ( (2 ± δ) + 2) \qquad\qquad\text{ eq. 2.1x-5}$$
Now remembering that $δ$ is always greater than zero (and more importantly it never is zero), we can see that in the above, we still never have to evaluate $f(x)$ at the forbidden value. So we just multiply out the above expression:
$$f(2 ± δ) = 8 ± 2 δ \qquad\qquad\text{ eq. 2.1x-6}$$
The requirement is that we have to be able to choose $δ$ so that the value above is no farther from the limit (which in this case is $8$) than the $ε$ that you might give me, no matter how small an $ε$ you do give me. So by giving me an $ε$, you are telling me to make it so that:
$$ |f(2 ± δ) - 8| ≤ ε \qquad\qquad\text{eq. 2.1x-7}$$
But we can get an expression for what's inside the absolute value brackets from stuff we have already done. Just take equation 2.1x-6 and subtract $8$ from both sides. If you substitute that in for $2 ± δ$, you get:
$$|± 2 δ| ≤ ε \qquad\qquad\text{eq. 2.1x-8}$$
and since the absolute value brackets make the ± sign moot, we have simply:
$$ 2 δ ≤ ε \qquad\qquad\text{ eq. 2.1x-9}$$
or
$$δ ≤ \frac{ε}{2} \qquad\qquad\text{eq. 2.1x-10}$$
So, if you hit me with any $ε$, all I have to tell you is to try a $δ$ that is less than or equal to half of your $ε$. In other words we have established a scheme that turns $ε$'s into $δ$'s, and the scheme always gives you a $δ$ that makes the function come within $ε$ of $8$. And that means we're done with the proof.
