﻿ Resolution of limits with zero indetermination between zero and zero

# Resolution of limits with zero indetermination between zero and zero. Exercises resolved.

I am going to explain to you now, with several exercises, how to solve the limits with indeterminate zero divided by zero.

In each of them, although the same procedure will be followed, you will see that each one has its nuances, so I recommend that you analyze each one of them carefully.

We begin by describing the general procedure for resolving limits with 0/0 indetermination.

## Procedure to solve limits with zero indetermination between zero

First of all, it should be noted that we do not know whether the limit will be determined or undetermined and, if it is undetermined, we do not know what indeterminacy it will be.

Therefore, the first step to solve any limit is to replace the x with the number you shop for and see what you get.

Let us suppose that after replacing and operating we reach the result 0/0, which is an indeterminacy.

From this point on, to resolve the indeterminations of the zero rate between zero and zero, the following procedure must be followed:

1. The polynomials of the numerator and denominator are broken down into factors.
2. We replace polynomials at the limit by their decomposition into factors.
3. The factors that are repeated in the numerator and in the denominator are eliminated. This eliminates indeterminacy
4. Replace the x with the number you are tending to, arriving at a certain solution.

The biggest difficulty of this procedure lies in the decomposition of polynomials into factors, so you must be very clear about how to decompose polynomials, as well as how to master the remarkable products, how to extract common factor, the Ruffini method…

## Resolved exercises of limits with zero indetermination between zero

Let’s solve a few step-by-step examples of limits with 0/0 indetermination so you can learn how to solve them.

This time, I am going to focus on the limit resolution procedure, without going into too much detail at each step, so that you have an overview above all.

Let’s go with the first example: First, we replace the x with the 3 to solve the limit and it gives us as a result the zero indeterminacy between zero and zero: Therefore, I am going to break down the polynomials of the numerator and the denominator into factors. The polynomial of the numerator is a remarkable product, so its decomposition is: The polynomial of the denominator cannot be broken down, as it is already grade 1 and is therefore already reduced to the maximum, so it remains the same.

I replace the polynomial of the numerator by its decomposition into factors and it remains: The factor (x-3) is repeated in the numerator and in the denominator so I can eliminate it: Looking like this: Once we have eliminated the repeated factors, the indeterminacy has also been eliminated, so we can replace the x with the 3 and reach the limit solution: Let’s solve another example: We replace the x with the 2 and operate. We’ve reached indeterminacy 0/0: In this case, both polynomials are grade 2, so they can be broken down into factors. We’ll break them down and we’ll have some left:  We replace polynomials by their decomposition into factors: And we eliminate the factors that are repeated in the numerator and in the denominator, which in this case is the factor (x-2): When we eliminate it, it’s left to us: We replace the x with the 2 and operate again. The indeterminacy has disappeared and we have reached the final result: Let’s solve one last example: We replace the x with the 0 and operate. The result is zero indetermination between zero and zero: We now break down the polynomials into factors. Sometimes, it is not necessary to break down the polynomial into grade 1 or irreducible factors. What we are trying to achieve by breaking them down is to find a factor that is repeated up and down to eliminate it and that the indeterminacy disappears.

So, this time, we see that we can draw common factor to the x in both polynomials:  And therefore, it is the x the factor that we eliminate in the numerator and in the denominator, so it is not necessary to continue decomposing the factor of grade 3, which has remained in the denominator:  When we eliminate the x of the numerator and the denominator we have left: That we replace the x with zero again and arrive at the final result: As I was telling you at the beginning, although the resolution procedure is the same, each limit has its details and you have to be prepared with a good mathematical basis to be able to break down any polynomial, since it is the difference between them.

## How to solve limits with zero indetermination between zero and root indetermination

Many times, limits with 0/0 indetermination have roots and in these cases it is not possible to factor the polynomials to eliminate the same factor from the numerator and the denominator.

How do you solve a limit with 0/0 indetermination that has roots?

In this case, the numerator and denominator must be multiplied by the conjugate of the binomial where the root is.

For example: We replace the x with the 1 and it gives us the result of zero indeterminacy between zero and zero: In this case, the root is in the denominator, therefore, of this binomial will be the conjugate by which we will have to multiply the numerator and the denominator: The denominator is a sum per difference, which is equal to a difference of squares: And in the numerator, we have a grade 2 factor that is a difference of squares, which we can break down as a sum by difference: And in this way, the factor (x-1) can be eliminated from the numerator and denominator: And we’re left with it: Now we can replace the x with the 1 and we have reached the solution, since we have eliminated the indeterminacy: Let’s see another example: We replace the x with the 3 and arrive at the result of zero between zero: We multiply the numerator and the denominator by the conjugate of the numerator’s binomial: In the numerator we have a sum per difference, which is equal to a difference of squares: In the denominator we have the binomial (x²-9), which is a difference of squares and we can put it as a sum per difference: By having it this way, we can eliminate the factor (x-3): And we’re left with it: We replace the x with the 3 and we reach the solution of the limit: 