Concept of limit in math: Definition, formula, Applications, and Examples

Concept of limit in math

Ancient Greek mathematician Archimedes of Syracuse (287–212 BC) gave the idea of limits first developed to calculate curved figures along with the volume of a sphere. To shape these figures into small pieces that can be calculated, after the number of pieces will increase, and also the limit of the sum of all pieces provide the required quantity.

English physicist and mathematician Sir Isaac Newton (1642–1727) and German mathematician Gottfried Wilhelm Leibniz (1646–1716) developed the general principles of calculus through their own efforts in the 17^th century.

In this article, we’ll explain the basic definition with the formula and important application of limits and examples.

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What is limit?

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

In mathematics, the concept of limit commonly shows the notion of arbitrary closeness. That is, a limit is a value that a variable quantity approaches as closely as one desire. The operations of derivatives and integral from calculus both depend on the theory of limits.

The theory of limits is based on a useful characteristic of real numbers; namely that between any two real numbers, no matter how close together they are, there is always another one. There are infinitely many more real numbers between two real numbers.

Formula of limit

The limit of a function is basically written as

Lim_x→c f(x) = L

And read as “the limit “f of x” as “x” approaches “c” equals to L”.

f(x) → L as x → c,

which reads “f of x tends to L as x tends to c”.

Applications of limit

The concept of limit is requiring to understand the real number system and it knowing the working. The basic idea about real numbers is real numbers can be defined as the numbers that are the limits of convergent sequences of rational numbers.

Another application of the concept of limits is on the derivative. The derivative is a rate of flow that can be evaluated depending on some limits concepts. Limits also play a vital role in evaluating integrals.

Limits are also part of the iterative process iteration is basically multiple times performing the same process by using the output of one step as an input of the next step and each time performing is called an iterate after a few successful iterates give you as close as required to an exact value.

Examples of limit

Example 1:

Determine the limit of 2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x, where x approaches 2.

Solution:

Step 1: Putting limits in the given function.

f(x) = 2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x

c = 2

Lim_x→c f(x) = Lim_x→2 (2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x)

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

Lim_x→2 [2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x)] = Lim_x→2 [2x⁵] – Lim_x→2 [6x⁴] + Lim_x→2 [11x²] + Lim_x→2 [9x] – Lim_x→2 [14] / Lim_x→2 [3x]

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

Lim_x→2 [2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x)] = 2Lim_x→2 [x⁵] – 6Lim_x→2 [x⁴] + 11Lim_x→2 [x²] + 9Lim_x→2 [x] – Lim_x→2 [14] / 3Lim_x→2 [x]

Step 4: Replacing limit value with x.

Lim_x→2 [2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x)] = 2(2⁵) – 6 (2⁴) + 11 (2²) + 9 (2) / 3 (2)

Lim_x→2 [2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x)] = 2 (2 * 2 * 2 * 2 * 2) – 6 (2 * 2* 2* 2) + 11 (2 * 2) + 18 – 14 / 6

Lim_x→2 [2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x)] = 2 (32) – 6 (16) + 11(4) + 18 – 14 / 6

Lim_x→2 [2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x)] = 64 – 96 + 44 + 18 – 14 / 6

Lim_x→2 [2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x)] = 64 – 96 + 44 + 18 – 14 / 6

Lim_x→2 [2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x)] = 126 – 96 – 14 / 6

Lim_x→2 [2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x)] = 30 – 14/6 = 30 – 7/3 = 90/3 – 7/3

Lim_x→2 [2x⁵ – 6x⁴ + 11x² + 9x – 14 / 3x)] = 83/3

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Example 2:

Determine the limit of the function f(x) = (x² – 9) / x – 3 where x approaches to 3.

Solution:

Step 1: Putting the limits in the given function

f(x) = (x² – 9) / x – 3

c = 3

 Lim_x→c F(x) = Lim_x→3 [(x² – 9) / (x – 3)]

Step 2: Now put the value of the limit.

Lim_x→3 [(x² – 9) / (x – 3)] = (3²) – 9 / 3 – 3

Lim_x→3 [(x² – 9) / (x – 3)] = 9 – 9 / 3 – 3

Lim_x→3 [(x² – 9) / (x – 3)] = 0 / 0

As it is clear that it is indeterminate on this limit so it should simplify first.

Step 3: Simplification

Lim_x→3 [(x² – 9) / (x – 3)] = Lim_x→3 [((x) ² – (3)²) / (x – 3)]

Lim_x→3 [(x² – 9) / (x – 3)] = Lim_x→3 [(x – 3) (x + 3) / (x – 3)]

Lim_x→3 [(x² – 9) / (x – 3)] = Lim_x→3 (x + 3)

Step 4: Now apply the limit.

Lim_x→3 [(x² – 9) / (x – 3)] = Lim_x→3 (x + 3)

Lim_x→3 [(x² – 9) / (x – 3)] = (3 + 3)

Lim_x→3 [(x² – 9) / (x – 3)] = 6

Summary

In this article, you have gone through the basics of limits definition with formula its application and history also given in the start and lastly for a better understanding example section is provided which helps to solve limit problems step-by-step explanation is given after reading and understanding this article anyone can easily defend this topic.