# Limit Calculus: Definition, Types, Properties, and Calculations

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In mathematics, the limit is one of the vast applications of calculus that is used to define other limit applications in different scenarios. Such as limit calculus involved in differentiating functions, integrating functions to find the area under the curve, and checking the nature of the function.

Especially limit calculus is used to find the behavior of the function at a particular point. Derivative calculus and integral calculus use limits to find the rate of change and the area under the curve respectively. In this article, we’ll explain limit calculus with its types, properties, and examples with calculations.

## What is the limit of calculus?

In mathematics, the limit is the branch of calculus that deals with the value of a function that approaches as the index at a particular point. In calculus and mathematical analysis, the limit is very essential to define and find the derivative by the first principle method, the area under the curve by definite integral, and the behavior of the function whether it is continuous or discontinuous.

The general expression of limit calculus is:

Limta f(t) = M

Or

Ltta f(t) = M

Where,

The function f(t) approaches to M as “t” approaches “a”.

## Types of Limit Calculus

There are three known types of limit calculus such as: left-side limit, right-side limit, and two-sided limit. Let us discuss them briefly.

### Left Sided Limit

The limit value approaching from the left hand side is known as the left sided limit. It can be written as:

Limta f(t) = M

### Right Sided Limit

The limit value approaching from the right hand side is known as the right sided limit. It can be written as:

Limta+ f(t) = M

### Two Sided Limit

The limit value approaching from the both right & left hand sides is known as two sided limit. It will have existed when the left sided limit becomes equal to the right sided limit. It can be written as:

Limta+ f(t) = Limta f(t)

## Properties of Limit Calculus

There are various properties of limit calculus. Here we are going to discuss some of them.

Constant property: When the particular point of limit is applied to a constant function, then the function remains unchanged. The expression of a constant is:

Limta f(u) = f(u)

Power Property: The power property of limit calculus is used when the function with an exponent is given. The expression of the exponent function is:

Limta f(t)m = [Limta f(t)]m

Constant Function Property: This property of limit calculus is used when a constant coefficient is present with the independent variable of the function. According to this property, that constant coefficient will be taken outside the limit notation. The expression of a constant function is:

Limta K * f(t) = K * Limta f(t)

Sum property: When two or more terms in a function are present with the plus sign “+” among them, then the sum property of limit calculus is used. According to this property, the notation of limit will be applied to each function separately.

Limta [f(t) + g(t) + h(t)] = Limta [f(t)] + Limta [g(t)] + Limta [h(t)]

Difference property: When two or more terms in a function are present with the minus sign “-” among them, then the sum property of limit calculus is used. According to this property, the notation of limit will be applied to each function separately.

Limta [f(t) – g(t) – h(t)] = Limta [f(t)] Limta [g(t)] Limta [h(t)]

Product property: When two or more terms in a function are present with the multiply sign “x” among them, then the sum property of limit calculus is used. According to this property, the notation of limit will be applied to each function separately.

Limta [f(t) x g(t) x h(t)] = Limta [f(t)] x Limta [g(t)] x Limta [h(t)]

## How to calculate limit calculus problems?

The problems of limit calculus could be solved either with the help of an online calculator or manually. Let us take a few examples to understand how to calculate limit calculus problems manually or by using a limit calculator.

## Example 1:

Evaluate the limit of 5t3 – 4t2 + 2t * 12t4 / 14t2 when t approaches 2.

Solution

Manually

Step 1: Take the multi-sign function and apply the notation of limit along with the particular point.

Limt2 [5t3 – 4t2 + 2t * 12t4 / 14t2]

Step 2: Now use the sum, product, difference, and quotient properties of limit calculus to apply the notation of limit separately to each term.

Limt2 [5t3 – 4t2 + 2t * 12t4 / 14t2] = Limt2 [5t3] – Limt2 [4t2] + Limt2 [2t] * Limt2 [12t4] / Limt2 [14t2]

Step 3: Now constant function property of limit calculus.

Limt2 [5t3 – 4t2 + 2t * 12t4 / 14t2] = 5Limt2 [t3] – 4Limt2 [t2] + 2Limt2 [t] * 12Limt2 [t4] / 14Limt2 [t2]

Step 4: Now use the power property of limit calculus to apply the particular point.

Limt2 [5t3 – 4t2 + 2t * 12t4 / 14t2] = 5 [23] – 4 [22] + 2 [2] * 12 [24] / 14 [22]

Limt2 [5t3 – 4t2 + 2t * 12t4 / 14t2] = 5 [8] – 4 [4] + 2 [2] * 12 [16] / 14 [4]

Limt2 [5t3 – 4t2 + 2t * 12t4 / 14t2] = 40 – 16 + 4 * 192 / 56

Limt2 [5t3 – 4t2 + 2t * 12t4 / 14t2] = 40 – 16 + 768 / 56

Limt2 [5t3 – 4t2 + 2t * 12t4 / 14t2] = 40 – 16 + 13.71

Limt2 [5t3 – 4t2 + 2t * 12t4 / 14t2] = 24 + 13.71

Limt2 [5t3 – 4t2 + 2t * 12t4 / 14t2] = 37.71

By using the limit calculator

Here are the steps to solve the above limit problem with the help of a limit calculator with steps.

Step 1: Enter the function.

Step 2: Write the particular point.

Step 3: Select the variable.

Step 4: Click submit button.

Step 5: The solution will show below calculate the button press to show more for the step by step solution of the problem.

## Example 2:

Evaluate the limit of 5u2 + 6u3 – 12u * 2 – 6u when u approaches 1.

### Solution

Manually

Step 1: Take the multi-sign function and apply the notation of limit along with the particular point.

Limt1 [5u2 + 6u3 – 12u * 2 – 6u]

Step 2: Now use the sum, product, and difference properties of limit calculus to apply the notation of limit separately to each term.

Limt1 [5u2 + 6u3 – 12u * 2 – 6u] = Limt1 [5u2] + Limt1 [6u3] – Limt1 [12u] * Limt1 [2] – Limt1 [6u]

Step 3: Now constant function property of limit calculus.

Limt1 [5u2 + 6u3 – 12u * 2 – 6u] = 5Limt1 [u2] + 6Limt1 [u3] – 12Limt1 [u] * Limt1 [2] – 6Limt1 [u]

Step 4: Now use power and constant properties of limit calculus to apply the particular point.

Limt1 [5u2 + 6u3 – 12u * 2 – 6u] = 5 [12] + 6 [13] – 12 [1] * [2] – 6 [1]

Limt1 [5u2 + 6u3 – 12u * 2 – 6u] = 5 [1] + 6 [1] – 12 [1] * [2] – 6 [1]

Limt1 [5u2 + 6u3 – 12u * 2 – 6u] = 5 + 6 – 12 * 2 – 6

Limt1 [5u2 + 6u3 – 12u * 2 – 6u] = 5 + 6 – 24 – 6

Limt1 [5u2 + 6u3 – 12u * 2 – 6u] = 11 – 24 – 6

Limt1 [5u2 + 6u3 – 12u * 2 – 6u] = –13 – 6

Limt1 [5u2 + 6u3 – 12u * 2 – 6u] = –19

## Conclusion

The limit is the most essential branch of calculus that has several types and properties to deal with the behavior of the function. The other calculus branches like derivative, integral, and continuity could be defined with the help of limit calculus.

Mayank Tewari