Limit is one of the most important and widely used concepts of calculus. It is a fundamental and core dimension of calculus that plays a crucial role in the study of integral calculus, differential calculus, series (Especially in the study of convergent series), continuous functions, etc.

As calculus developed in the **18 ^{th} century**, scientists most probably mathematicians treated the limit concept intuitively. Limit plays a key role in mathematical analysis like in the study of continuity of functions. The limit of a function exists if the outputs of a function get close to some value as its inputs get close to another value.

In this blog, we will explore the idea of limits, and its definition as well as we will discuss some important theorems of limits with solved examples.

Table of Contents Show

## Table of Contents

**Definition of Limit of a Function**

Let a function f(x) be defined in an open interval near a number “a” but not equal to that of “a”. If the given function f(x) approaches a particular number “L” as x (which is independent variable) approaches to “a” from both sides of “a” (from the left and right side), then “L” will be the limit of the given function f(x) as “x” approaches “a”. Mathematically,

Lim _{x🡪an} f(x) = L and it is read as “limit of f(x) is L as x → “a”.

It is not advisable and practicable to compute the limit of a function-making numerical approach. We compute the limits of the given functions in some mechanical way.

**Theorems of Limits**

Consider f(x) and g(x) are two functions for which

Lim_{x}_{🡪}_{a} f(x) = L and Lim_{x}_{🡪}_{a} g(x) = M, then there are some important theorems (rules) on limits of functions in the following table, that help out during the computation of limits of functions.

1. | Summation Theorem | Lim_{x🡪a} [ f(x) + g(x)] = Lim_{x🡪} f(x) + Lim_{x🡪a} g(x) = L + M |

2. | Difference Theorem | Lim_{x🡪a} [ f(x) – g(x) ] = Lim_{x🡪a} f(x) – Lim _{x🡪a }g(x) = L – M |

3. | Constant Function Theorem | Lim_{x🡪a} [ kf(x) ] = k Lim_{x🡪a} f(x) = kL |

4. | Product Theorem | Lim_{x🡪a} [ f(x) g(x) ] = [ Lim_{x🡪a} f(x) ] [ Lim_{x🡪a} g(x) ] = LM |

5. | Quotient Theorem | Lim_{x🡪a} [ f(x) / g(x) ] = [ Lim_{x🡪a} f(x) ] / [ Lim_{x🡪a} g(x) ] = L / M |

6. | Power Theorem | Lim_{x🡪a} [ f(x) ]^{n} = [ Lim_{x🡪a} f(x) ]^{n} = L^{n} |

**How to calculate limits?**

The limit theorems can be used for finding the limits of a function at a specific point manually. Alternatively, a **limit calculator** can be used to calculate limits with steps. Below are a few examples of limit calculus.

**Example 1: **

Compute the limit of the function (x + 5) at point 1.

**Solution: **

**Step 1:** Given the data

f(x) = x + 5, x = 1

**Step 2:** Apply limits on the given function and simplify.

Lim_{x🡪1} (x+ 5) = Lim_{x🡪1} x + Lim_{x🡪1} 5 (Summation theorem)

Lim_{x🡪1} (x+ 5) = 1 + 5 = 6 Ans.

So, **6** is the limit of the given function.

**Example 2: **

Find out the limit of the function (x – 5) at point 3.

**Solution:**

**Step 1:** Given the data

Function = x – 5, x = 3

**Step 2:** Apply limits on the given function and simplify.

Lim_{x🡪3} (x -5) = Lim_{x🡪3} x – Lim_{x🡪3} 5 (Difference theorem)

Lim_{x🡪3} (x -5) = 3 – 5 = -2 Ans.

So, **-2** is the limit of the given function.

**Example 3: **

Compute the limit of the function (7x) at point 3.

**Solution:**

**Step 1:** Given the data

Function = 7x, x = 3

Step 2: Apply the limits on the given function and simplify.

Lim_{x🡪3} (7x) = 7 Lim_{x🡪3} (x) (Constant function theorem)

Lim_{x🡪3} (7x) = 7 (3)

Lim_{x🡪3} (7x) = 21 Ans.

So, **21 **is the limit of the given function.

**Example 4:**

Compute the limit of the function (2x^{2}) (2x + 4) at point 1.

**Solution: **

**Step 1:** Given the data

Function = (2x^{2}) (2x + 4) and x = 1.

**Step 2:** Apply the limits on the given function and simplify.

Lim_{x🡪1} [(2x^{2}) (2x + 4)] = [Lim_{x🡪1} 2x^{2}] [ Lim_{x🡪1} (2x + 4)] (Product theorem)

Lim_{x🡪1} [(2x^{2}) (2x + 4)] = [2 ( Lim_{x🡪1} x^{2} )] [ Lim_{x🡪1} (2x) + Lim_{x->1} 4 ]

Lim_{x🡪1} [(2x^{2}) (2x + 4)] = [2 (-1)^{2} ] [2 (1) + 4 ]

Lim_{x🡪1} [(2x^{2}) (2x + 4)] = 2 (1) (2 + 4)

Lim_{x🡪1} [(2x^{2}) (2x + 4)] = 2 (6) = 12 Ans.

So, **12 **is the limit of the given function.

**Example 5: **

Find out the limit of the function (6x + 4) / (x + 2)

**Solution: **

**Step 1:** Given the data

Function = (3x +4) / (x + 3), x= 2

**Step 2:** Apply limits on the given function and simplify.

Lim_{x🡪2} [(3x +4) / (x + 3)] = [Lim_{x🡪2} (6x + 4)] / [ Lim_{x🡪2} (x + 2)] (Quotient theorem)

Lim_{x🡪2} [(3x +4)/ (x + 3)] = (6(2) + 4)/ (2 + 2)

Lim_{x🡪2} [(3x +4) / (x + 3)] = (12 + 4) / 4 = 16/4

Lim_{x🡪2} [(3x +4)/ (x + 3)] = 4 Ans.

So, **4 **is the limit of the given function.

**Example 6: **

Compute the limit of the function (3x – 1)^{2} at point 2.

**Solution:**

**Step 1:** Given the data

Function = (3x – 1)^{2}, x = 2

**Step 2:** Apply the limit on the given function and simplify.

Lim_{x🡪2} (3x – 1)^{2} = [Lim_{x🡪2} (3x – 1)]^{2} (Power theorem)

Lim_{x🡪2} (3x – 1)^{2} = [3(2) – 1]^{2}

Lim_{x🡪2} (3x – 1)^{2} = (6 – 1)^{2}

Lim_{x🡪2} (3x – 1)^{2} = (5)^{2} = 25

Hence **25** is the required answer.

**Summary**

In this article, we have discussed the core concept of limits. We described its definition as well as we briefed some important theorems on limits. In the last section, we solved some examples following the rules (theorems) of limits.