Find your limits

I'm sure by now most of you are more than familiar with the concept of limit of a function, but I believe it is a great place to start our blog with.

At first, the idea of limit of a function in a point was used by mathematicians without worrying of having a precise definition for it. When the time came for mathematical analysis principles to be formalized, it was A.L. Cauchy (1789-1857) who wrote the definitive version, the one we still use today. It goes like this:

A function f(x) has a limit L in a point p if for every real number ε > 0 exists another real number δ > 0 such as |f(x) - L| < ε if 0 < |x - p| < δ and x ≠ p, where δ is chosen depending on ε. It is written

But, what does this exactly mean? What this definition tells us is that when x takes values close to p, as close as you want, but without being equal to p, both from the left and the right of p (higher and lower values than p), the function approaches to a value L, which is the limit of the function, as much as we want.

With this definition it comes natural to assume that for the limit L to exist, it's value must be the same whether we approach from the left or from the right. In fact, this is the only requirement for the limit to exist, as we don't need the function to exist in a certain point for it to have a limit in that point. We will understand this better with an example:

We want to find the limit of $$\displaystyle\lim_{x \to{1}}{\displaystyle\frac{x-1}{x^2-1}}$$

If we try replacing the x by 1 we will get the following: 
$$ \frac{x-1}{x^{2}-1} = \frac{1-1}{1^{2}-1} = \frac{0}{0} $$
This is what we call an indetermitation, which means that the function does not have a representation in that point, ie, it does not exist.

In this case we have that f(1) does not exist, but we can still try to find the limit. If we take a closer look at how the function behaves when approaching 1 from right and left, it's relatively easy to see where things are going.

f(1,5)= 0,4...                                           f(0,5)= 0,66...
f(1,25)= 0,44...                                       f(0,75)= 0,57...
f(1,1)= 0,47...                                         f(0,9)= 0,52...
f(1,01)= 0,49...                                       f(0,99)= 0,502...

Right now it seems intuitive to think that if there was a f(1), it would be equal to 0,5. 
So, how can we solve this limit without having to make so many tedious calculations? Well, there are several tools, techniques and methods you can use to solve limits, but usually the simplest solution is the best one, let's see 
$$ \frac{x-1}{x^{2}-1} = \frac{x-1}{\left(x-1\right) \cdot \left(x+1\right)} = \frac{1}{x+1} = \frac{1}{2} $$
By cancelling out the (x-1) we get the solution to the limit.

This is problably the easiest method and the easiest type of limit, but math, as building houses, always has to start with the foundations. There will me more math to come.....stay tuned!!