# Teach and Learn Limits in Math Easily by Using an Online Limit Calculator

Teach and Learn Limits in Math: The limit of a function indicates whether the function is finite or infinite in its nature. The infinite function has no limits and it is directly solved by implementing specific calculations like derivation or integration. When we are solving a finite function, we can solve it by using the **limit calculator**. This methodology is suitable for the students, just to get the quick answer to the limit in practice. If we are solving a function, algebraically, you can solve a limit of a function by four techniques and we need to choose one of the techniques:

These techniques are of solving the limit of a function are as follows:

- The substitution method
- Factoring method
- Rationalizing the numerator method
- Lowest common denominator method

Now we start to learn the limits of the algebraic functions by the following methodologies, the limits calculator also makes our solution more reliable. We can check if the answer is correct by using the limit calculator.

We need to use different techniques as some of the limits are easily evaluated by the substitution, some are solved by factoring. The others are solved by rationalizing as the substitution and the factoring methods are not applicable here. When we are dealing with the complex rational number, then we are not able to use all three techniques, that is substitution, factorizing, and rationalization. Here, we are compelled towards the LCD, or the Least Common Multiple methods, to solve the limits.

Table of Contents

**The substitution method:**

The first technique we can use is by algebraically substituting the value of the “x” in the approaching function. If you find the undefined value of the, especially when the denominator is “0”. Then we need to move to another technique, we can also use the limit calculator with steps to solve the limits in steps. If your function does not have the values of the “0” at the denominator position. Then the function is continuous at all values of the “x”. Then by the substitution method, we can solve the function, then you can find the limit by the substitution method:

F(x)=** ****x****5****x****2****-6x+8****x-4**

Now, when we are putting the value of the limit=5, then we would get:

**(5)****2****-6(5)+8****5-4**= **25-30+8****5-4**= **33-30****1**

The answer to that limit world is the f(5)= 3 and we can conclude the limit is continuous. We can solve the limit by using the limit solver, which would make your answer more reliable.

**Factoring method:**

If the substitution methodology is not applicable, then we can use the factoring method. This method is more suitable when any part of the function is polynomial, now we want to solve the following expression. Then by the factoring method, we can solve it as follows:

F(x)=** ****x****4****x****2****-6x+8****x-4**

For example, if we substitute the 4, in the limit then, you would get “0” in the numerator. It means you can’t solve this function by the substitution method when putting the “4” in the limit. Then we go for the factoring methodology.

** **F(x)=** ****x****4****x****2****-6x+8****x-4**

Now by factoring, we would get the expression as:

** **F(x)=** ****x****4****x****2****-6x+8****x-4**= **x****4** **x****2****-4x-2x+8****x-4**= **x****4** **(x-4)(x-2)****x-4**

Now the (x-4), is in both the denominator and the numerator, and it cancels out its effects, and we only remain (x-2).

F(x)=** ****x****4**(x-2)= (4-2)

F(x)= 2

This is the answer to the limit, which can only be solved by the factoring method, not by the substitution method, the limit calculator can be the best to solve the limit by the factoring method.

**Rationalizing the numerator method:**

In rationalizing the limit algebraically, you need to rationalize the limit first, especially the numerator. These functions have the square roots in the numerator and the polynomial expression in the denominator. For example, you need to find the limit of this expression as it approaches “13”. Then :

F(x)=**x****13**x-4 -3x-13

Now when we put the “13” in the denominator, we get the “0”, which makes the fraction become impossible to solve.

- Now we make the conjugate, which is actually the conjugate of the numerator, and we then cancel it out by multiplying it by both the numerator and the denominator; we can also use the limit calculator in solving such limits.
- Multiplying the top and bottom of the function by the conjugate, which is x-4+3.
- Multiplying through by the conjugate, we would get:
- x-4 -3x-13.x-4+3x-4+3
- Now cancel the factors, we would get:

x-13(x-13).x-4 +3

- The term (x-13), each other effect, leaving the following expression:

1.x-4 +3

- Now put the limit in the function, we would finally get the answer, which is 1/6, and the solution of the limit. We can also solve the expression by the limit
**calculator**, for our convenience.

**Lowest common denominator method:**

When you are dealing with the complex rational function, then we normally use the LCD method to solve the limit. The substitution method fails here, as you end up with “0” in the denominator. The function is not factorizable, and you have no square root to make the conjugate. So we need to move towards the last technique, which is the LCD method, the limit calculator can be used to solve such difficult limits:

Now we use the technique of LCD on the following function:

Consider the function6

F(x)= x01 x+6x–16

Now by LCD

Now by solving the limit by the LCD method, we would get the following answer: which is -1/36, when the x, approaches “0”.

**The last word:**

The limit calculator is the best tool to find the limit of a function. As in some cases, we can’t solve the limit by some methods like Substitution, **factorization**. So we use the rationalization method, but in the case of complex rational numbers, we are using the LCD method. In this case, the limit is not solvable by any of the three above methods.