Limits of functions

`{\lim}\limits_{x \to a}f(x)`

The previous expression is read as Limit of the function f(x) as x approaches a.

Definition of Limit

This definition was created in the 19th century by the French mathematician Augustin Louis Cauchy and later refined to its current form by Karl Weierstrass.

For the expression `{\lim}\limits_{x \to c}f(x)=L`, it means that:

For every ε > 0, there exists δ > 0 such that for every x from 0 < |x − c| < δ, it holds that |ƒ(x) − L| < ε

Mathematically written as:

The Simplest Limits
Among the simplest limits would be, for example, `{\lim}\limits_{x \to 2}x+1`. When solving this example, we take the number that x approaches and substitute it into the function. It would look something like this: `{\lim}\limits_{x \to 2}x+1=2+1=3`. This limit was very easy because the function was defined at the point x=2. It gets trickier with the example `{\lim}\limits_{x \to 2}\frac{(x-2)(x+1)}{x-2}`. If you try to substitute x = 2 now, you will get zero in the denominator, which is obviously not possible. Therefore, we need to somehow eliminate the term (x-2), which in this case is very simple; we simply cancel out the entire expression: `{\lim}\limits_{x \to 2}\frac{(x-2)(x+1)}{x-2}={\lim}\limits_{x \to 2}x+1`. Now we can substitute 2 for x, and we conclude that as x approaches 2, the limit of the function is 3.
Solve the example `{\lim}\limits_{x \to 2}\frac{x^2-5x+6}{x^2-3x+2}`. The first step should be to try substituting the number two for x. However, if we do this, we find that we get zero in the denominator, and so we must try to simplify the expression somehow. 
`{\lim}\limits_{x \to 2}\frac{x^2-5x+6}{x^2-3x+2}`
Both the numerator and the denominator can be factored
`{\lim}\limits_{x \to 2}\frac{x^2-5x+6}{x^2-3x+2}={\lim}\limits_{x \to 2}\frac{(x-3)(x-2)}{(x-2)(x-1)}`
We can cancel
`{\lim}\limits_{x \to 2}\frac{x^2-5x+6}{x^2-3x+2}={\lim}\limits_{x \to 2}\frac{(x-3)(x-2)}{(x-2)(x-1)}=\frac{x-3}{x-1}`
Now we can substitute x = 2 without any problems
`{\lim}\limits_{x \to 2}\frac{x^2-5x+6}{x^2-3x+2}={\lim}\limits_{x \to 2}\frac{(x-3)(x-2)}{(x-2)(x-1)}=\frac{x-3}{x-1}=\frac{-1}{1}=-1`

Solve `{\lim}\limits_{x \to -2}\frac{x^4-16}{x^3+8}`
`{\lim}\limits_{x \to -2}\frac{x^4-16}{x^3+8}`
Both the numerator and the denominator can be factored
`{\lim}\limits_{x \to -2}\frac{x^4-16}{x^3+8}={\lim}\limits_{x \to -2}\frac{(x^2+4)(x^2-4)}{(x+2)(x^2-2x+4)}`
We further factor the numerator
`{\lim}\limits_{x \to -2}\frac{x^4-16}{x^3+8}={\lim}\limits_{x \to -2}\frac{(x^2+4)(x^2-4)}{(x+2)(x^2-2x+4)}={\lim}\limits_{x \to -2}\frac{(x^2+4)(x+2)(x-2)}{(x+2)(x^2-2x+4)}`
We can cancel
`{\lim}\limits_{x \to -2}\frac{x^4-16}{x^3+8}={\lim}\limits_{x \to -2}\frac{(x^2+4)(x^2-4)}{(x+2)(x^2-2x+4)}={\lim}\limits_{x \to -2}\frac{(x^2+4)(x+2)(x-2)}{(x+2)(x^2-2x+4)}={\lim}\limits_{x \to -2}\frac{x^2+4}{x^2-2x+4}`
Now we can substitute
`{\lim}\limits_{x \to -2}\frac{x^4-16}{x^3+8}={\lim}\limits_{x \to -2}\frac{(x^2+4)(x^2-4)}{(x+2)(x^2-2x+4)}={\lim}\limits_{x \to -2}\frac{(x^2+4)(x+2)(x-2)}{(x+2)(x^2-2x+4)}={\lim}\limits_{x \to -2}\frac{x^2+4}{x^2-2x+4}=-\frac{8}{3}`


General Approach to Solving Limits
When solving limits of functions, we follow these steps:

First, try to substitute the number that the function approaches into the limit. In most cases, you will get an indeterminate form, but it’s worth a try.
Try to simplify the expression → factor something and then cancel it out. In the example `{\lim}\limits_{x \to 2}\frac{x^2-5x+6}{x^2-3x+2}`, x approaches the number 2. It is therefore clear that in the denominator, we need to eliminate the term `(x-2)` → `{\lim}\limits_{x \to 2}\frac{x^2-5x+6}{x^2-3x+2}={\lim}\limits_{x \to 2}\frac{(x-3)(x-2)}{(x-2)(x-1)}=\frac{x-3}{x-1}=-1`
If you can’t cancel anything out, you can try using L'Hôpital's rule (but more on that later).


Rules for Calculating Limits
The following rules are useful, especially when dealing with complex limits. 

`{\lim}\limits_{x \to a}\left[f(x)+g(x)\right]={\lim}\limits_{x \to a}f(x)+{\lim}\limits_{x \to a}g(x)`
`{\lim}\limits_{x \to a}\left[f(x)-g(x)\right]={\lim}\limits_{x \to a}f(x)-{\lim}\limits_{x \to a}g(x)`
`{\lim}\limits_{x \to a}\left[f(x)\cdot g(x)\right]={\lim}\limits_{x \to a}f(x)\cdot{\lim}\limits_{x \to a}g(x)`
`{\lim}\limits_{x \to a}\left[\frac{f(x)}{g(x)}\right]=\frac{{\lim}\limits_{x \to a}f(x)}{{\lim}\limits_{x \to a}g(x)}`
`{\lim}\limits_{x \to a}c=c` (where c is any constant)