Limit is a branch of calculus (a branch of mathematics that deals with the study of properties and behavior of functions) that is a fundamental concept used to understand the nature/behavior of the function at a particular point.

This type of calculus plays a vital role in defining other types of calculus such as integration, differentiation, continuity, etc. In this article, we are going to explain limit calculus along with its types and calculations.

## Limit Calculus

In calculus, the limit value of a function at a specific point is a fundamental concept that allows us to understand the behavior of the function as it gets closer to the particular point. The basics of this type of calculus is to find the behavior of the function as it approaches a specific point, as there is no concern with the actual value of the function.

Mathematically, the limit of a function “h(v)” as the independent variable “v” approaches “d” equals to P such as:

**Lim _{v}**

_{→d}**[h(v)] = P**

Where

- “Lim” = the notation of limit
- “v→d” = represents v approaches specific point “d”
- “h(v)” = given function that is under consideration
- “P” = limit value of the function at “d”

## What is meant by approaching a value?

Let us understand the concept of approaching a value with the help of a general example of a function h(v) = v^{2} as the independent variable “v” approaches “2”.

At v = 1.5

Lim_{v}_{→1.5} [v^{2}] = 1.5^{2} = 2.25

At v = 1.9

Lim_{v}_{→1.9} [v^{2}] = 1.9^{2} = 3.61

At v = 1.99

Lim_{v}_{→1.99} [v^{2}] = 1.99^{2} = 3.9601

At v = 1.999

Lim_{v}_{→1.999} [v^{2}] = 1.999^{2} = 3.996001

At v = 2

Lim_{v}_{→2} [v^{2}] = 2^{2} = 4

At v = 2.001

Lim_{v}_{→2.001} [v^{2}] = 2.001^{2} = 4.004001

At v = 2.01

Lim_{v}_{→2.01} [v^{2}] = 2.01^{2} = 4.0401

At v = 2.1

Lim_{v}_{→2.1} [v^{2}] = 2.1^{2} = 4.41

As the independent variable “v” goes closer to 2 the function h(v) goes closer to 4.

## Types of Limit Calculus

There are several types of limit calculus depending on the behavior of the function and the approaching point.

### One-sided Limit

In limit calculus, when a function “h(v)” approaches a particular point from only one side is said to be the one-sided limit. The one-sided limit could be left-sided or right-sided. When the function’s approaching value comes from the left side, then it will be named the left-sided limit.

The left-sided limit is denoted as:

Lim_{v}_{→d^-} [h(v)]

When the function’s approaching value comes from the right side, then it will be named a right-sided limit.

The right-sided limit is denoted as:

Lim_{v}_{→d^+} [h(v)]

### Two Sided Limit

In limit calculus, when a function “h(v)” approaches a particular point from both left and right sides is said to be the two-sided limit. When both the left-sided limit and the right-sided limit existed, then a two-sided limit will exist.

Lim_{v}_{→d} [h(v)] = Lim_{v}_{→d^-} [h(v)] = Lim_{v}_{→d^+} [h(v)]

If the left-sided limit and right-sided limit are not equal or one of them not existed, then the two-sided limit does not exist.

## Laws of Limit Calculus

Laws of limit calculus are a set of rules that help us to evaluate the limit value of various algebraic functions at a particular point. Below are the main laws of limit calculus.

### 1. Sum Law

The law of sum is applied to two or more sum of functions. According to this law, the limit will be applied to each function individually. The mathematical expression of this law is:

Lim_{v}_{→d} [h(v) + g(v)] = Lim_{v}_{→d} [h(v)] + Lim_{v}_{→d} [g(v)]

### 2. Difference Law

The law of difference is applied to two or more difference of functions. According to this law, the limit will be applied to each function individually. The mathematical expression of this law is:

Lim_{v}_{→d} [h(v) – g(v)] = Lim_{v}_{→d} [h(v)] – Lim_{v}_{→d} [g(v)]

### 3. Product Law

The law of product is applied to two or more product of functions. According to this law, the limit will be applied to each function individually. The mathematical expression of this law is:

Lim_{v}_{→d} [h(v) x g(v)] = Lim_{v}_{→d} [h(v)] x Lim_{v}_{→d} [g(v)]

### 4. Quotient Law

The law of quotient is applied to two or more divisible functions. According to this law, the limit will be applied to each function individually. The mathematical expression of this law is:

Lim_{v}_{→d} [h(v) / g(v)] = Lim_{v}_{→d} [h(v)] / Lim_{v}_{→d} [g(v)]

### 5. Power Law

The law of power is applied to exponential functions. According to this law, the limit will be applied first. The mathematical expression of this law is:

Lim_{v}_{→d} [h(v)]^{p} = [Lim_{v}_{→d} h(v)]^{p}

## How to Solve the Problems of Limit Calculus?

You can use a limit calculator to solve limit problems with step-by-step solutions in no time. Let us solve some examples to understand how to evaluate the problems of limit calculus manually.

**Example 1:**

Evaluate the limit of h(v) = 2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3 if the independent variable “v” approaches “3”.

**Solution**

**Step 1:** Write the given function according to the mathematical representation of the limit.

h(v) = 2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3

Lim_{v}_{→}_{d} [h(v)] = Lim_{v}_{→3} [2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3]

**Step 2:** Now use the laws of product, sum, and difference to make the function simple.

Lim_{v}_{→3} [2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3] = Lim_{v}_{→3} [2v^{3}] x Lim_{v}_{→3} [12v^{2}] – Lim_{v}_{→3} [2v^{4}] + Lim_{v}_{→3} [15v] + Lim_{v}_{→3} [v^{5}] – Lim_{v}_{→3} [3]

**Step 3:** Now take out the constant terms outside the notation of limit.

Lim_{v}_{→3} [2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3] = 2Lim_{v}_{→3} [v^{3}] x 12Lim_{v}_{→3} [v^{2}] – 2Lim_{v}_{→3} [v^{4}] + 15Lim_{v}_{→3} [v] + Lim_{v}_{→3} [v^{5}] – Lim_{v}_{→3} [3]

**Step 4:** Now use the direct substitution method and substitute t = 4 in the above expression.

Lim_{v}_{→3} [2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3] = 2 [3^{3}] x 12 [3^{2}] – 2 [3^{4}] + 15 [3] + [3^{5}] – [3]

Lim_{v}_{→3} [2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3] = 2 [27] x 12 [9] – 2 [81] + 15 [3] + [243] – [3]

Lim_{v}_{→3} [2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3] = 54 x 108 – 162 + 45 + 243 – 3

Lim_{v}_{→3} [2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3] = 5832 – 162 + 45 + 243 – 3

Lim_{v}_{→3} [2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3] = 5670 + 45 + 243 – 3

Lim_{v}_{→3} [2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3] = 5715 + 243 – 3

Lim_{v}_{→3} [2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3] = 5958 – 3

Lim_{v}_{→3} [2v^{3} x 12v^{2} – 2v^{4} + 15v + v^{5} – 3] = 5955

**Example 2**

Evaluate the limit as “v” approaches 9 of the function h(v) = (√v – 3) / (v – 9)

**Solution**

**Step 1:** Write the given function according to the mathematical representation of the limit.

h(v) = (√v – 3) / (v – 9)

Lim_{v}_{→}_{d} [h(d)] = Lim_{v}_{→9} [(√v – 3) / (v – 9)]

**Step 2:** Now use the direct substitution method and substitute t = 3 in the above expression.

Lim_{v}_{→9} [(√v – 3) / (v – 9)] = Lim_{v}_{→9} [(√v – 3)] / Lim_{v}_{→9} [(v – 9)]

Lim_{v}_{→9} [(√v – 3) / (v – 9)] = (Lim_{v}_{→9} [√v] – Lim_{v}_{→9} [3]) / (Lim_{v}_{→9} [v] – Lim_{v}_{→9} [9])

Lim_{v}_{→9} [(√v – 3) / (v – 9)] = ([√9] – 3) / ([9] – [9])

Lim_{v}_{→9} [(√v – 3) / (v – 9)] = (3 – 3) / (9 – 9)

Lim_{v}_{→9} [(√v – 3) / (v – 9)] = 0/0

This is the indeterminate form of the function. we have to use the rationalization method for removing the indeterminate form of function.

**Step 3:** Now rationalize the given function

√v – 3) / (v – 9) = [(√v – 3) ÷ (v – 9)] x [(√v + 3) ÷ (√v + 3)]

√v – 3) / (v – 9) = [(√v – 3) x (√v + 3)] ÷ [(v – 9) x (√v + 3)]

√v – 3) / (v – 9) = [(v – 9)] ÷ [(v – 9) x (√v + 3)]

**Step 4:** Take the given function, place the factors, and cancel the common factors.

√v – 3) / (v – 9) = ~~[(v – 9)]~~ ÷ [~~(v – 9) ~~x (√v + 3)]

√v – 3) / (v – 9) = 1 ÷ (√v + 3)]

**Step 5:** Now apply the limits again

Lim_{v}_{→9} [(√v – 3) / (v – 9)] = Lim_{v}_{→9} [1] ÷ Lim_{v}_{→9} [ (√v + 3)]

Lim_{v}_{→9} [(√v – 3) / (v – 9)] = Lim_{v}_{→9} [1] ÷ (Lim_{v}_{→9} [√v] + Lim_{v}_{→9} [3])

Lim_{v}_{→9} [(√v – 3) / (v – 9)] = [1] ÷ ([√9] + [3])

Lim_{v}_{→9} [(√v – 3) / (v – 9)] = 1 ÷ (3 + 3)

Lim_{v}_{→9} [(√v – 3) / (v – 9)] = 1 ÷ 6 or 1/6

## Conclusion

Now you can take assistance in solving problems of limit calculus from this post as we have discussed the laws and solved examples. The basics of limit calculus have been covered in this post to make it easy to understand.