To calculate the limit of a function it is important to bear in mind the basic properties of the limits.

## Properties of the limits

Let $$f(x)$$ be a function, therefore:

- $$\displaystyle\lim_{x \to p}{x}=p \\ $$
- $$\displaystyle\lim_{x \to p}{k\cdot f(x)}=k\cdot\lim_{x \to p}{f(x)}$$ where $$k$$ is an integer.
- $$\displaystyle\lim_{x \to p}{(f(x)+g(x))}=\lim_{x \to p}{f(x)}+\lim_{x \to p}{g(x)} \\ $$
- $$\displaystyle\lim_{x \to p}{(f(x)-g(x))}=\lim_{x \to p}{f(x)}-\lim_{x \to p}{g(x)} \\ $$
- $$\displaystyle\lim_{x \to p}{(f(x)\cdot g(x))}=\lim_{x \to p}{f(x)}\cdot \lim_{x \to p}{g(x)} \\ $$
- $$\displaystyle\lim_{x \to p}{\dfrac{f(x)}{g(x)}}=\dfrac{\displaystyle\lim_{x \to p}{f(x)}}{\displaystyle\lim_{x \to p}{g(x)}}$$ si $$g(x)\neq0$$

Here we can see an example of each property:

- $$\displaystyle\lim_{x \to 2}{x}=2 \\ $$
- $$\displaystyle\lim_{x \to 1}{2x^2}=2\cdot\lim_{x \to 1}{x^2}=2\cdot1^2=2 \\ $$
- $$\displaystyle\lim_{x \to 4}{(x^2+3x)}=\lim_{x \to 4}{x^2}+\lim_{x \to 4}{3x}=4^2+3\cdot4=16+12=28 \\ $$
- $$\displaystyle\lim_{x \to 3}{(x-5x)}=\lim_{x \to 3}{x}-\lim_{x \to 3}{5x}=3-5\cdot3=3-15=-12 \\ $$
- $$\displaystyle\lim_{x \to 2}{x^2}=\lim_{x \to 2}{x}\cdot \lim_{x \to 2}{x}=2\cdot2=4 \\ $$
- $$\displaystyle\lim_{x \to 1}{\dfrac{x+1}{x}}=\dfrac{\displaystyle\lim_{x \to 1}{x+1}}{\displaystyle\lim_{x \to 1}{x}}=\dfrac{1+1}{1}=2$$

## Operations with infinity

We know that some limits can turn out to be infinite. This fact and the previous properties can lead us to situations where we will have to add up, remain, multiply and divide by the value of infinity.

$$$\lim_{x \to +\infty}{(x-x^2)}=\lim_{x \to +\infty}{x}-\lim_{x \to +\infty}{x^2}=\infty-\infty$$$

Is the value of this operation zero? What infinity is bigger, the one of $$x$$ or the one of $$x$$ square? This situation is said to be an indetermination.

Indetermination can be of many types:

$$$\infty-\infty,\dfrac{\infty}{\infty},\dfrac{0}{\infty},1^\infty,\ldots$$$

You can learn how to solve these situations in the following unit (calculation of limits-indeterminations).

Now, we are able to calculate the limit of a function.

To compute a limit of a function $$f(x)$$ in $$x=p$$ we simply have to substitute $$x$$ for $$p$$ in the expression of the limit. When doing this step, a concrete value (a number, an infinite value or an indetermination can be obtained. In the last case we will have to modify the expression of the limit to another equivalent expression in order to aempty the indetermination. This step is explained in the following unit, (calculation of limits-indeterminations).

In the case where functions are defined by parts or non-continuous functions we will have to do side limits depending on the point at which the limit has to be computed. We will choose the suitable expression of the function to set in the limit if we approach from the right or from the left. We have an example:

Consider $$f(x)=\left\{\begin{array}{c} x+1 \ \text{ si } x < 1 \\ x-1 \ \text{ si } x\geq1 \end{array} \right.$$ and we'll look the side limits in $$x=1$$.

$$$L^-=\lim_{x \to 1^-}{f(x)}=\lim_{x \to 1^-}{x+1}=1+1=2$$$

$$$L^+=\lim_{x \to 1^+}{f(x)}=\lim_{x \to 1^+}{x-1}=1-1=0$$$

When we do a limit of a function in infinity it is necessary to proceed in the same way as if we were doing the limit at a point, but replacing the value of the concrete point with infinity.

We will have to distinguish the values of "plus infinity" or "minus immensely" when doing the limit and we will obtain the result. When we do the limit, what can occur is that we obtain a finite value, an infinite value or an indetermination.

We will also have to be careful when using the signs at the time of doing the limit.

If we compute the limit of the cubic function when $$x$$ tends to minus infinity:

$$$\lim_{x \to -\infty}{x^3}=\lim_{x \to -\infty}{x}\cdot \lim_{x \to -\infty}{x}\cdot\lim_{x \to -\infty}{x}=(-\infty)\cdot(-\infty)\cdot(-\infty)=-\infty$$$

while if we realize the limit of the quadratic function:

$$$\lim_{x \to -\infty}{x^2}=\lim_{x \to -\infty}{x}\cdot \lim_{x \to -\infty}{x}=(-\infty)\cdot(-\infty)=+\infty$$$