# Limit in Calculus: Defined and Explained With Its Rules and Examples

Limits are an important part of calculus, and they play an important role in many different areas of mathematics and physics. The concept of a limit in calculus is often confusing for students. In this section, we’ll provide a definition of limit in calculus as well as the rules.

We’ll also explain how to calculate limits and discuss the importance of limits in calculus.

## What Is the Limit In Calculus?

A limit is a value a function approaches as the input gets closer and closer to some value. When we say that something is limited, what we’re really saying is that there is a point at which the quantity involved will stop increasing or decreasing.

Limu→a [f(u)] = L

In other words, there is a point at which it will hit its maximum or minimum value. In terms of practical applications, Limits play an important role in many fields of mathematics and science including calculus, physics, engineering design, and economics.

Without understanding limits, it would be difficult to solve problems or design products that meet the required specifications. For example, when engineers are designing a bridge or airplane they need to know how strong the structure will be under various conditions (say wind speed and weight).

Without knowing how strong the structure will be under various conditions (at different points along its length), it would be impossible to make an informed decision about which designs to choose for testing purposes.

## Laws Of Limit In Calculus

The Limit Laws are used when the limit can be calculated using basic algebraic operations – such as addition, subtraction, multiplication, and division – without having to factor or cancel out any parentheses. These laws are as follows:

• ### Constant Law

When there is a constant coefficient or any algebraic expression given without a corresponding variable, then the constant Law of limit calculus is used. According to this law, the limit of a constant function remains unchanged at any point.

Limu→a [C] = C

• ### Constant function Law

When there are constant coefficients along with the corresponding variable, then constant function law is used. According to this law, the constant coefficient will be taken outside the limit notation.

Limu→a [C f(u)] = C Limu→a [f(u)]

• ### Sum Law

If the algebraic expression is given along with the plus sign among them, then the sum law of limit calculus is used. According to this law, the notation of limit will apply to each term of the algebraic expression separately.

Limu→a [f(u) + g(u)] = Limu→a [f(u)] + Limu→a [g(u)]

• ### Difference Law

If the algebraic expression is given along with the minus sign among them, then the difference law of limit calculus is used. According to this law, the notation of limit will apply to each term of the algebraic expression separately.

Limu→a [f(u) – g(u)] = Limu→a [f(u)] – Limu→a [g(u)]

• ### L’hopital’s Law

L’hopital’s law of limit calculus is used when a function makes an indeterminate form such as 0/0 or infinity/infinity. According to this law, you have to find the derivative of the numerator and denominator and then apply the limit value.

Limu→a [f(u) / g(u)] = Limu→a [d/du f(u) / d/du g(u)]

• ### Product Law

If the algebraic expression is given along with the multiply sign among them, then the product law of limit calculus is used. According to this law, the notation of limit will apply to each term of the algebraic expression separately.

Limu→a [f(u) * g(u)] = Limu→a [f(u)] * Limu→a [g(u)]

• ### Quotient Law

If the algebraic expression is given along with the division sign among them, then the quotient law of limit calculus is used. According to this law, the notation of limit will apply to each term of the algebraic expression separately.

## How to calculate limits by using its rules?

To calculate the problems of the limits, you have to follow the below examples. You can also take assistance from a limit solver to get the step-by-step solution to the given problems.

Example 1

Find the limit of f(u) = 5u – 6u3 + 4u * 2u2, as the specific point is 4.

Solution

Step 1: Take the given algebraic expression and apply the notation of limit calculus to it.

Limu→a [f(u)] = Limu→a [5u – 6u3 + 4u * 2u2]

Step 2: Now use the sum, difference, and product laws of limit calculus.

Limu→a [5u – 6u3 + 4u * 2u2] = Limu→a [5u] – Limu→a [6u3] + Limu→a [4u] * Limu→a [2u2]

Step 3: Now apply the constant function law.

Limu→a [5u – 6u3 + 4u * 2u2] = 5Limu→a [u] – 6Limu→a [u3] + 4Limu→a [u] * 2Limu→a [u2]

Step 4: Substitute the specific point to the above expression.

Limu→4 [5u – 6u3 + 4u * 2u2] = 5 [4] – 6 [43] + 4 [4] * 2 [42]

Limu→4 [5u – 6u3 + 4u * 2u2] = 5 [4] – 6 [4 x 4 x 4] + 4 [4] * 2 [4 x 4]

Limu→4 [5u – 6u3 + 4u * 2u2] = 5 [4] – 6 [64] + 4 [4] * 2 [16]

Limu→4 [5u – 6u3 + 4u * 2u2] = 20 – 384 + 16 * 32

Limu→4 [5u – 6u3 + 4u * 2u2] = 20 – 384 + 512

Limu→4 [5u – 6u3 + 4u * 2u2] = – 364 + 512

Limu→4 [5u – 6u3 + 4u * 2u2] = 148

Example 2

Find the limit of f(w) = (2w2 – 18) / (w – 3) as the specific point is 3.

Solution

Step 1: Take the given algebraic expression and apply the notation of limit calculus to it.

Limw→a [f(w)] = Limw→3 [(2w2 – 18) / (w – 3)]

Step 2: Now use the sum, difference, and product laws of limit calculus.

Limw→3 [(2w2 – 18) / (w – 3)] = (Limw→3 [2w2] – Limw→3 [18]) / (Limw→3 [w] – Limw→3 [3])

Step 3: Now apply the constant function law.

Limw→3 [(2w2 – 18) / (w – 3)] = (2Limw→3 [w2] – Limw→3 [18]) / (Limw→3 [w] – Limw→3 [3])

Step 4: Substitute the specific point to the above expression.

Limw→3 [(2w2 – 18) / (w – 3)] = (2 [32] – [18]) / ([3] – [3])

Limw→3 [(2w2 – 18) / (w – 3)] = (2 [3 x 3] – [18]) / ([3] – [3])

Limw→3 [(2w2 – 18) / (w – 3)] = (2 [9] – [18]) / ([3] – [3])

Limw→3 [(2w2 – 18) / (w – 3)] = (18 – 18) / (3 – 3)

Limw→3 [(2w2 – 18) / (w – 3)] = 0/0

Step 5: Now use L’hopital’s law of limit calculus as the given function makes an indeterminate form.

Limw→3 [(2w2 – 18) / (w – 3)] = Limw→3 [d/dw (2w2 – 18) / d/dw (w – 3)]

Limw→3 [(2w2 – 18) / (w – 3)] = Limw→3 [d/dw (2w2) – d/dw (18) / d/dw (w) – d/dw (3)]

Limw→3 [(2w2 – 18) / (w – 3)] = Limw→3 [(4w) – (0) / (1) – (0)]

Limw→3 [(2w2 – 18) / (w – 3)] = Limw→3 [(4w) / (1)]

Limw→3 [(2w2 – 18) / (w – 3)] = Limw→3 [(4w)]

Apply the specific point again.

Limw→3 [(2w2 – 18) / (w – 3)] = 4(3)

Limw→3 [(2w2 – 18) / (w – 3)] = 12

## Wrap Up

The concept of a limit is very important in calculus because it allows us to find the behavior of a function as it approaches a certain point. By understanding the rules of calculating limits, we can better understand how functions behave near specific points. The examples in this blog have hopefully given you a better understanding of limits and how they work in calculus.