The limit functions are related to calculus and analysis. It is a value that a function approaches as that function’s inputs get closer and closer to some other number. Derivatives and definite integrals are defined based on limit functions. Mathematically,

The limit of f(x) as x approaches x_{0} is L, i.e.

f(x) = L

If, for every∈ > 0, there exists δ >0 such that, for all x.

Index

**Explanation**

Limits tell us where the specific function is going.

Let us consider the function

F(x) = 3x^{2} -1

According to this function, we will first see that what value it is approaching. Let us use the x value close to the number 4. Let’s see by using some inputs.

X | F(X) |

3.9 | 46.63 |

3.99 | 46.7603 |

3.999 | 46.976003 |

For the above-written values, we will say that function is approaching 47. So you can say that the limit of the function as x approaches 4 is 47. There is one thing to remember that we can change to indiscriminately close to the number 47, just have to pick the values that are sufficiently close to 4 to do it. but here a new question arises, could not you have found the limit by plugging 4 into the function? Would not that also give you 47? Yes, in this case, the focus with the limit should be on the value its attempting to achieve, and there are some functions where you can’t plug in a number to find the limit. To put it another way, they are not necessarily the same.

Consider a different scenario

F(x) = x^{2}-4/x-2

We are interested in what value the function is trying to reach, just as we are previously. We use the value of x close to 3.

x | F(X) |

2.9 | 4.9 |

2.99 | 4.99 |

2.999 | 4.999 |

So again we say the limit of the function as x approaches 3, is 5. If you try and find this value by instead plugging in 3, something strange happens. When you plug 3 into the function, you will receive a result of 00. This shows that the function does not actually ever get to 5. In fact, a quick look at the graph shows a hole right at 5. Despite the hole, the behaviour of the function leading up to it is the same. Because the comportment is the same, we continue to assert that the function’s maximum as x approaches 3 is 5, even if never reaches it.

**Methods to calculate limits**

There are a lot of methods of calculating the limits. Some of them are listed below.

- By plugging in the “ x “ value.
- By using a calculator device.
- Rationalizing the numerator.
- Finding the lowermost similar denominator.
- By factoring.
- By using an online limit calculator.

**Calculating limits with examples**

#### Example 1

Let’s say we are given the following limit function to work with.

x-6x+8/x-4

We will first plug 4 into the function as a result you will get zero in the numerator and denominator. The quadratic expression in the numerator will tell you to factorize it by mid-term breaking. The equation will make the following factor (x-2)(x-4). As (x-4) is present in the numerator and denominator so they will cancel out each other.

We will be left with: F(x) = x – 2

But it has a hole when x = 4 because the original function is still undefined there (because it creates 0 in the denominator). However, if the term in the denominator did not cancel and the value is still undefined, the limit of the function at the value of x does not exist

Now let us consider another example

F(x) = x^{2}-3x-28/x^{2}-6x-7

= (x-7)(x+4)/(x-7)(X+1)

Now if you asked the limit of the function as x approaches 7, you could connect 7 into the cancellation sort and get 11/8 .

**Example 2 (By Plugin Method)**

Suppose we are given the following limit

x^{2}-2x-3/x^{2}-9

=(4)^{2}-2(4)-3/(4)^{2}-9

=5/7

So the simple technique is, first you need to see where x is approaching. In the given example x is approaching 4. So you have to put or plug 4 wherever you find x. which gave you the approximate result of the limit functions. But the actual problem comes when you’ve left with zero in the denominator.

**Example 3 (Factorization):**

So let us suppose that we are given the following limit function

x^{2}-2x-3/x^{2}-9

If we apply the plugin method we will be left with the zero in the denominator. You will not only get zero in the denominator but also in the numerator which is a complete indeterminate form. So this expression is begging you to factorize it.

(x-3)(x+1)/(x+3)(x-3)

=x+1/x+3

=3+1/3+3

=2/3

The point of factoring was that it got rid of these problem terms ( x -3)/(x -3) that term that was creating 0/0 form. Now that they have dropped out, you can plug in and just get the actual number that the limit is equal to.

**The limit does not exist (DNE)-condition:**

Let’s assume we are given the following problem to work with.

x^{2}+2x-8/x^{2}+5x+4

Look at this one, What if you’ve factored, cancelled term. You plug in again at the end, and you still get zero in the denominator. Which is undefined. This means that your function is increasing without limit shooting off to infinity or negative infinity. It is limitless, and you can write it as DNE (does not exist).

**By Common Denominator:**

Look at the following example. Here you can neither factorize nor plugin. So you just have to make the denominator the same to get the required result.

**Open up parenthesis**

(x+2)^{2}-4x

Here is another common example. It is also very useful. First, you have to try the plugin. It will be left with an indeterminate form after solving. You can neither factorize nor take the common denominator as the present condition shows. Here is the fourth approach, see if you can expand everything, multiply out, distribute, and then simplify. Hopefully, something cancels and you will be able to the plugin, in the end, and find the required limit. The above-written expression can also be written after the simplification.

(x+4)

= 0+4=4

## Finding limit by using online Limit Calculator:

There are multiple online calculators for finding the limit functions. The online limit calculator can help you to evaluate the specific limit.

You will see the following screen.

Here is just a demo, you and fill out your function and your value to approach. Now, you just have to click calculate. And the answer will be shown exactly at the bottom. You can also see the step-by-step method if you are unable to understand how the answer came. You can also have access to specify the variable, specify the direction, and including the second limit.

**Note: the value of a continuous function at a given location is its limit**

**Applications**

- Measuring the temperature is a limit.
- Limits are used to calculate the derivatives as a daily life problem.
- They help us to make an exact quantity of an approximation.

**Conclusion**

The limit function is the basic part of calculus mathematics. Hope you have grasped all the strategies for determining the problem of the limits.