If the whole idea of limits has seemed a little too vague, then we're in luck. It's time to pull ourselves out of the calculus confusion. Let's don our prom attire and prepare for the formal, precise definition for limit.
Definition. Let f(x) be a function. We say the limit of f(x) as x approaches a is L, written
if for every real number ε > 0 there exists a real number δ > 0 such that
if |x – a| < δ then |f(x) – L| < δ.
This will make a lot more sense with a picture. Here's the idea: we're claiming that as x "gets closer" to a, f(x) "gets closer" to L, therefore the function would look something like this:
or perhaps like this, with a hole in the graph:
If we try to translate this formal definition into English, it's saying "no matter how close we want f(x) to be to L, if we pick x to be close enough to a we'll find the solution."
The real number ε is how close we want f(x) to be to L. That is, we want all the function values to fall within the shaded band:
The real number δ is how close x needs to be to a for the function values to fall within the shaded band:
If x is anywhere between a – δ and a + δ, then f(x) will fall between L – ε and L + ε.
Sample Problem
Let f(x) = 2x. Then
If we take, as an example, ε = 0.5, how close does x need to be to 1 in order for f(x) = 2x to be within 0.5 of 2?
Saying we want f(x) to be within 0.5 of 2 means we want f(x) to be in between 1.5 and 2.5 (in the shaded band on the above graph).
For f(x) to be greater than 1.5 we need x to be greater than 0.75, and for f(x) to be less than 2.5 we need x to be less than 1.25. Drawing this on the graph, we see that if x is within 0.25 of 1 then f(x) will be in the shaded band.
If we took any other value of ε > 0, we could find some other value of δ > 0 so that |x – 1| < δ would guarantee |f(x) – 2| < ε. For this function, as long as 
, we're in the clear. Therefore the limit of f(x) as x approaches 1 is equal to 2.
Sample Problem
Let 
This function looks like this:
It's tempting to say that
However, if we take a tiny value of ε, no matter how small we take δ to be, there will be some troublesome values of f(x). By troublesome we mean points where x is within δ of 1, but f(x) is not within ε of 5. Therefore the limit of f(x) as x approaches 1 can't possibly be 5.
No matter how small we take δ to be, there will always be some x-values between 1 – δ and δ with f(x) = 3. If ε is anything less than 2, then f(x) won't be within ε of 5.
That's just how the cookie crumbles.