# Definition, Application, and Examples of Limit Calculus

In Calculus, the concept of limit is defined as a value that a function approaches the output for the given input values. it plays a vital role in

In Calculus, the concept of limit is defined as a value that a function approaches the output for the given input values. it plays a vital role in calculus and calculation; it is used to determine to define integrals and derivatives. It addresses how the function operates at a certain time and is employed in the starting process.

The notion of limit in topology is more extensive and relates to the concepts of direct limit and limit in theory. There are two further types to categorize the integrals indefinite and definite integrals limits are defined and while indefinite integrals limits are not shown.

In this article, the basic definition of limit its formula, and detailed with the help of examples will be discussed easily.

## Define limit

Limit is a value that a function (or sequence) approaches as the input (or index) approaches some value”

## Formula:

\lim_{x \rightarrow c}\ f(x)=L

## Applications

The real number system must be understood, along with how it operates, to fully grasp the concept of limit. The fundamental concept behind real numbers is that they may be described as the maximum or minimum of convergent sequences of rational numbers.

The derivative is a further area in which constraints are applied. The derivative is a flow rate that can be assessed using various limit theories. Limits are crucial in the assessment of integrals.

The iterative method includes limitations as well. Iteration is the practice of doing a task more than once while using the results of one step as the input for a subsequent one. Many successful iterations will give you a value that is as near to being exact as is necessary.

## Methods to evaluate limits

In this section, we will discuss the different methods which help to evaluate limits. Each method will be used in examples according to their use. Moreover, according to the situation of the problem, we will use suitable methods to get their answers.

1. Substitution method
2. Factorization
3. Rationalization
4. Reduction to Standard form
5. L’ Hospital Rule

## Example Section:

In this section with the help of an example, the topic is explained.

Example 1:

What is the limit of function (3x5 – 4x4 + 5x2 + 6x – 1) / 2x, where x approaches 2?

Solution:

Step 1: Using the limits in a given function

f(x) = (3x5 – 4x4 + 5x2 + 6x – 1)/ 2x

c = 2

Limxc f(x) = Limx2 (3x5 – 4x4 + 5x2 + 6x – 1)/ 2x

Step 2: Applying the limits notation separately by using the rules of the limits mentioned in the above table.

Limx f(x) = Limx2 (3x5 – 4x4 + 5x2 + 6x – 1 / 2x) = {Limx2 (3x5) – Limx2 (4x4) + Limx2 (5x2) + Limx2 (6x) – Limx2 (1)} / Limx2 (2x)

Step 3: Write the constants outside the limit using the constant multiplication rule.

Limx2 (3x5 – 4x4 + 5x2 + 6x – 1 / 2x) = 3 Limx2 (x5) – 4 Limx2 (x4) + 5 Limx2 (x2) + 6 Limx2 (x) – Limx2 (1) / {2 Limx2 (x)}

Step 4: Changing the limit value with the variable x.

Limx2 (3x5 – 4x4 + 5x2 + 6x – 1 / 2x) = 3(25) – 4 (24) + 5 (22) + 6 (2) / 2 (2)

Limx2 (3x5 4x4 + 5x2 + 6x 1 / 2x) = {3 (2 * 2 * 2 * 2 * 2) 4 (2 * 2* 2* 2) + 5 (2 * 2) + 12} / 4

Limx2 (3x5 4x4 + 5x2 + 6x 1 / 2x) = {3 (32) 4 (16) + 5(4) + 12} / 4

Limx2 (3x5 4x4 + 5x2 + 6x 1 / 2x) = 96 + 64 + 20 + 12 / 4

Limx2 (3x5 4x4 + 5x2 + 6x 1 / 2x) = 160 + 20 + 12 / 4

Limx2 (3x5 4x4 + 5x2 + 6x 1 / 2x) = 192/ 4

Limx2 (3x5 4x4 + 5x2 + 6x 1 / 2x) = 192 / 4

Limx2 {(3x5 4x4 + 5x2 + 6x 1 / 2x} = 48

A limit solver can be used to calculate the above problem online without involving into larger calculations.

Example 2:

What is the limit of the function f(x) = (x3 - 27) / 2x – 6 where x approaches 3.

Solution:

Step 1: Using the limits in the given function

f(x) = (x3 - 27) / 2x – 6

c = 3

Limxc f(x) = Limx3 {(x3 - 27) / 2x – 6)}

Step 2: Now put the value of the limit.

Limx3 {(x3 - 27) / 2x 6)} = (33) 27 / 2(3) 6

Limx3 {(x3 - 27) / 2x 6)} = 27 27 / 6 6

Limx3 {(x3 - 27) / 2x 6)} = 0 / 0

It is an indeterminate form

Step 3: With the help, of the L’ Hospital rule by taking the derivative of the numerator and denominator make it in the defined form

Limx3 {(x3 - 27) / 2x 6)} = Limx3 d/dx (x3-27) / d/dx (2x - 6)

Limx3 {(x3 - 27) / 2x 6)} = Limx3 3x2 / 2

Step 4: Now changing variable x with limit 3

Limx→3 {(x3 - 27) / 2x – 6)} = 3(3)2/2

Limx3 {(x3 - 27) / 2x – 6)} = 27 / 2

## Summary:

In this article, the basic definition of limit its formula, and different methods are discussed which helps to evaluate the limits of different functions. Moreover, through examples topic is explained and for the detailed explanation, their applications are also discussed.