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.

~~(toc)~~

## 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 2x^{5}
– 6x^{4} + 11x^{2} + 9x – 14 / 3x, where x approaches 2.

**Solution:**

**Step 1:** Putting limits in the given function.

f(x)
= 2x^{5} – 6x^{4} + 11x^{2} + 9x – 14 / 3x

c
= 2

Lim_{x}_{→}_{c} f(x)
= Lim_{x}_{→}_{2}
(2x^{5} – 6x^{4} + 11x^{2} + 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} [2x^{5}
– 6x^{4} + 11x^{2} + 9x – 14 / 3x)] = Lim_{x}_{→}_{2} [2x^{5}]
– Lim_{x}_{→}_{2} [6x^{4}]
+ Lim_{x}_{→}_{2}
[11x^{2}] + 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} [2x^{5}
– 6x^{4} + 11x^{2} + 9x – 14 / 3x)] = 2Lim_{x}_{→}_{2} [x^{5}]
– 6Lim_{x}_{→}_{2} [x^{4}]
+ 11Lim_{x}_{→}_{2}
[x^{2}] + 9Lim_{x}_{→}_{2} [x]
– Lim_{x}_{→}_{2} [14]
/ 3Lim_{x}_{→}_{2} [x]

**Step 4:** Replacing limit value with x.

Lim_{x}_{→}_{2} [2x^{5}
– 6x^{4} + 11x^{2} + 9x – 14 / 3x)] = 2(2^{5}) – 6 (2^{4})
+ 11 (2^{2}) + 9 (2) / 3 (2)

Lim_{x}_{→}_{2} [2x^{5}
– 6x^{4} + 11x^{2} + 9x – 14 / 3x)] = 2 (2 * 2 * 2 * 2 * 2) – 6
(2 * 2* 2* 2) + 11 (2 * 2) + 18 – 14 / 6

Lim_{x}_{→}_{2} [2x^{5}
– 6x^{4} + 11x^{2} + 9x – 14 / 3x)] = 2 (32) – 6 (16) + 11(4) +
18 – 14 / 6

Lim_{x}_{→}_{2} [2x^{5}
– 6x^{4} + 11x^{2} + 9x – 14 / 3x)] = 64 – 96 + 44 + 18 – 14 /
6

Lim_{x}_{→}_{2} [2x^{5}
– 6x^{4} + 11x^{2} + 9x – 14 / 3x)] = 64 – 96 + 44 + 18 – 14 /
6

Lim_{x}_{→}_{2} [2x^{5}
– 6x^{4} + 11x^{2} + 9x – 14 / 3x)] = 126 – 96 – 14 / 6

Lim_{x}_{→}_{2} [2x^{5}
– 6x^{4} + 11x^{2} + 9x – 14 / 3x)] = 30 – 14/6 = 30 – 7/3 =
90/3 – 7/3

**Lim _{x}**

_{→}

_{2}**[2x**

^{5}– 6x^{4}+ 11x^{2}+ 9x – 14 / 3x)] = 83/3Alternatively you can take assistance through online
tools to save time.

*Solved through **limitcalculator.online*

**Example
2:**

Determine
the limit of the function f(x) = (x^{2} – 9) / x – 3 where x approaches to 3.

**Solution:
**

**Step 1:** Putting the limits in the given function

f(x)
= (x^{2} – 9) / x – 3

c
= 3

Lim_{x}_{→}_{c} F(x) = Lim_{x}_{→}_{3} [(x^{2}
– 9) / (x – 3)]

**Step 2:** Now put the value of the limit.

Lim_{x}_{→}_{3} [(x^{2}
– 9) / (x – 3)] = (3^{2}) – 9 / 3 – 3

Lim_{x}_{→}_{3} [(x^{2}
– 9) / (x – 3)] = 9 – 9 / 3 – 3

Lim_{x}_{→}_{3} [(x^{2}
– 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} [(x^{2}
– 9) / (x – 3)] = Lim_{x}_{→}_{3} [((x)^{
2} – (3)^{2}) / (x – 3)]

Lim_{x}_{→}_{3} [(x^{2}
– 9) / (x – 3)] = Lim_{x}_{→}_{3} [(x
– 3) (x + 3) / (x – 3)]

Lim_{x}_{→}_{3} [(x^{2}
– 9) / (x – 3)] = Lim_{x}_{→}_{3} (x
+ 3)

**Step 4:** Now apply the
limit.

Lim_{x}_{→}_{3} [(x^{2}
– 9) / (x – 3)] = Lim_{x}_{→}_{3} (x
+ 3)

Lim_{x}_{→}_{3} [(x^{2}
– 9) / (x – 3)] = (3 + 3)

Lim_{x}_{→}_{3} [(x^{2}
– 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.

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