# Limits Definition, Formulas & Examples |Calculus|

There are some mathematical function for which you cannot find the right solution for certain values.

For Example:
In the function, f(x) = \frac{x^{2} -1}{x-1}
you cannot find the value of function when x = 1, as the calculation will be:
\Longrightarrow \frac{1^{2} -1}{1-1}\\\ \\ \Longrightarrow \frac{1}{0}\\\ \\ \Longrightarrow \infty
Hence the value is indeterminate and we are now stuck with no right answer.

## Introduction to Limits

Through limits we can get approximate value of function f(x), when x=1.
In this approach we will try to find the value of function when the value of x approaches towards 1.

In the above table you can observe that when x approaches towards 1, the value of f(x) is moving closer towards number 2. This statement can be represented as:

\lim _{x\rightarrow 1} \ \frac{x^{2} -1}{x-1} \ =2 \\\ \\

It says that when x approaches towards 1, the limit of the function is equal to 2.

Note: Limit does not gives you the actual value of the function. It just provides you with approximate value for the given point.

### What is Limit?

Limit tells the value of function around certain point, instead at the point.
We use the concept of limit when we can’t calculate the value of function at a point, so instead we try to find its approximate value near that point.

Limit is represented as:

From above illustration one can write as:
“As x approaches 1, the limit of function f(x) is c” or
The approximate value of function f(x), when x approaches 1 is c

### Right Hand Limit and Left Hand Limit

There are two ways to find the approximate value of the function
(1) Right hand limit : we find the value from right of x
(2) Left hand limit : we find the value from left of x

#### Right Hand and Left Hand Limits Example

Example 01
Find the value of following function, when x =2
\frac{x^{2} -4}{x-2}

Solution
When we put x=2 in the equation we get,
\Longrightarrow \frac{2^{2} -4}{2-2}\\\ \\ \Longrightarrow \ \frac{4-4}{0}\\\ \\ \Longrightarrow \infty

Hence , we can’t determine the value of function using this method.

(a) Left Hand Limit
let us now try to find the approximate value of function when the value of x approaches from 1 to 2

You can observe from the above table that as the value of x approaches towards 2, the value of function is approaching towards 4.
Hence we can say that:
\lim _{x\rightarrow 2}\frac{x^{2} -4}{x-2} \ =4\

This is the left hand limit of the given function

(b) Right Hand Limit
Let us now find the approximate value of function from right side when x=4

From the table, you can observe that as the value of x approaches towards 2, the value of function move towards number 4.

\lim _{x\rightarrow 2}\frac{x^{2} -4}{x-2} \ =4\

Hence the right hand limit of the function is also 4

Graphical Representation

Example 02
f(x) = x + 10
Find the limit of function at x = 5

Note:
Limits can be calculated for simple function also.
Even if we know the answer of the expression, we can calculate the limits

Solution
(a) Left hand Limit

As the value of x approaches 5, the value of f(x) approach towards 15.

Hence, the limit of function can be written as:
\lim _{x\rightarrow 5}\ x+10\ =15\

(b) Right Hand Limit

You can observe from the above table that as the value of x approaches closer to 5, the value of f(x) moves closer to 15. Hence the limit can be represented as:

\lim _{x\rightarrow 5}\ x+10\ =15\

From both the right and left hand limit, we found that the approximate limit value of function for x = 5 is 15

Example 03
f(x) = x + cos x
Find the limit of function when x approaches 0
(i.e. \lim _{x\rightarrow 0} \ \ x+cosx\ )

Solution
(a) Left Hand Limit

From the table you can observe that as the value of x approaches from -0.3 to 0, the value of function gets closer to 1. So we can say that:
\lim _{x\rightarrow 0} \ \ x+cosx\ =\ 1

(b) Right Hand Limit

From the table you can observe that as the value of x approaches from 0.3 to 0, the value of function f(x) moves closer to 1. Again we can say that:
\lim _{x\rightarrow 0} \ \ x+cosx\ =\ 1

Graphical Representation

Example 04
Find the limit of function when x approaches 0
f(x) = 1, x\leqslant 0
2, x > 0

Solution
This function can graphically represented as

Left Hand Limit of Function
\lim _{x\rightarrow 0} \ \ f( x) \ =2
When we approach the function from left of 0, we get value 1

Right Hand Limit of Function
\lim _{x\rightarrow 0} \ \ f( x) \ =2
Here if we approach the function from right of 0, we get 2

Hence we get different value of limit from different approach.
So the limit of function f(x) when x approach to 0 does not exist.

Example 05
Find the limit of function f(x) = sin(x)
When x approaches \pi /2

Solution
Left Hand Limit

From the table you can observe that as the value of x approaches \pi /2 , the value of function f(x) reaches 1.
\lim _{x\rightarrow \frac{\pi }{2}} \ \ sin( x) \ =1

Right Hand Limit

You can observe from the table that as the value of x approaches \pi /2 , the value of function approaches 1.
\lim _{x\rightarrow \frac{\pi }{2}} \ \ sin( x) \ =1

## Properties of Limits

(01) Limit of sum property

\lim_ {x\rightarrow a} \ [ f( x) +g( x)] \ =\lim_ {x\rightarrow a} \ [ f( x)] \ +\lim _{x\rightarrow a} \ [ g( x)]

(02) Limit of difference property

\lim_ {x\rightarrow a} \ [ f( x) \ - \ g( x)] \ =\lim_{x\rightarrow a} \ [ f( x)] -\lim _{x\rightarrow a} \ [ g( x)]

(03) Limit of product property

\lim_ {x\rightarrow a} \ [ f( x) \ \times \ g( x)] \ =\lim_ {x\rightarrow a} \ [ f( x)] \ \times \ \ \lim _{x\rightarrow a} \ [ g( x)]

(04) Limit with Constant

\lim_{x\rightarrow a} \ [ k\times f( x)] \ =k\times \lim_{x\rightarrow a} \ [ f( x)] \

(05) Limit of Division Property

\lim_{x\rightarrow a} \ \left[\frac{f( x)}{g( x)} \ \right] \ =\ \frac{\lim_{x\rightarrow a} \ [ f( x)]}{\lim _{x\rightarrow a} \ [ g( x)]} \ \

### Limit Theorem

\lim _{x\rightarrow a} \ \left[\frac{x^{n} -a^{n}}{x-a} \ \right] \ =\ na^{n-1} \

### Limit Trigonometric Theorem

(a) \lim _{x\rightarrow 0} \ \left[\frac{sin\ x}{x} \ \right] \ =\ 1\\\ \\

(b) \lim _{x\rightarrow 0} \ \left[\frac{1-cosx}{x} \ \right] \ =\ 0\

## Limit Questions using Formulas

(01) Find the limit of following function
\lim _{x\rightarrow 2} \ \left[\frac{x^{3} -4x^{2} +4x}{x^{2} -4} \ \right] \

Solution
If we put x=2 in the given function:
\Longrightarrow \ \left[\frac{2^{3} -4\times 2^{2} +4\times 2}{2^{2} -4} \ \right] \\\ \\ \Longrightarrow \ \ \left[\frac{2^{3} -4\times 2^{2} +4\times 2}{4-4} \ \right] \\\ \\ \Longrightarrow \ \ \left[\frac{2^{3} -4\times 2^{2} +4\times 2}{0} \ \right]\

The value is indeterminate here.
Let us try to rewrite the equation so that we can cancel some of the factors in the function.

\Longrightarrow \ \left[\frac{x^{3} -4x^{2} +4x}{x^{2} -4} \ \right] \\\ \\ \Longrightarrow \ \ \left[\frac{x( x-2)^{2}}{( x-2)( x+2)} \ \right] \\\ \\ \Longrightarrow \ \ \left[\frac{x\ ( x-2)}{( x+2)} \ \right] \\\ \\ Putting\ x=2\ in\ the\ above\ equation,\ we\ get\ \\\ \\ \Longrightarrow \ \left[\frac{2\ ( 2-2)}{( 2+2)} \ \right]\ \\\ \\ \Longrightarrow \ \frac{0}{4}\\\ \\ \Longrightarrow \ 0

Hence, 0 is the limit for the given function

(02) Find the limit of the following function
\lim _{x\rightarrow 2} \ \left[\frac{x^{3} -2x^{2}}{x^{2} -5x+6} \ \right] \

Solution
If we put x=2 in the equation, we get
\Longrightarrow \left[\frac{2^{3} -2\times 2^{2}}{2^{2} -5\times 2+6} \ \right]\\\ \\ \Longrightarrow \ \left[\frac{2^{3} -2\times 2^{2}}{10-10} \ \right] \\\ \\ \Longrightarrow \left[\frac{2^{3} -2\times 2^{2}}{0} \ \right]

Here we get 0 at the denominator, which means the value is indeterminate in this method.
Let us try to simplify the equation and cancel some of the factors

\Longrightarrow \left[\frac{x^{3} -2x^{2}}{x^{2} -5x+6} \ \right]\\\ \\ \Longrightarrow \left[\frac{x^{2}( x-2)}{( x-2) \ ( x-3)} \ \right]\\\ \\ \Longrightarrow \left[\frac{x^{2}}{( x-3)}\right] \\\ \\ putting\ x=2\\\ \\ \Longrightarrow \left[\frac{4}{2-3}\right]\\\ \\ \Longrightarrow -1

(03) Find the limit of following function
\lim _{x\rightarrow 1} \ \left[\frac{x^{15} -1}{x^{10} -1} \ \right]

Solution
Multiplying and dividing the limit with (x-1)
The limit is then written as:

\Longrightarrow \ \lim {x\rightarrow 1} \ \left[\frac{x^{15} -1}{x^{10} -1} \ \times \frac{x-1}{x-1}\right]\\\ \\ \Longrightarrow \ \lim {x\rightarrow 1} \ \left[\frac{x^{15} -1}{x-1} \ \times \frac{x-1}{x^{10} -1}\right]\\\ \\ Using\ the\ formula\\\ \\ \lim _{x\rightarrow a} \ \left[\frac{x^{n} -a^{n}}{x -a} \ \right] \ =\ n\ a^{n-1}\\\ \\ \Longrightarrow \ 15\ \times \frac{1}{10}\\\ \\ \Longrightarrow \ 1.5\

(04) Find the limit of following expression
\ \lim _{x\rightarrow 0} \ \left[\frac{tan( x)}{x} \ \right]\

Solution
We\ know\ that:\\\ \\ tan\ x\ =\ \frac{sin\ x}{cos\ x}\\\ \\ \ Putting\ the\ value\ in\ equation\ we\ get\\\ \\ \ \Longrightarrow \ \lim_{x\rightarrow 0} \ \left[\frac{sin( x)}{x} \times \frac{1}{cos( x)} \ \right] \ -\ -\ -eq( 01)\\\ \\ We\ know\ the\ formula\\\ \\ \lim_{x\rightarrow 0} \ \left[\frac{sin( x)}{x}\right] \ =1\\\ \\ \ Using\ this\ formula\ in\ the\ equation\ 01\\\ \\ \Longrightarrow \lim_{x\rightarrow 0} \ \left[\frac{sin( x)}{x}\right] \times \lim _{x\rightarrow 0}\left[\frac{1}{cos( x)} \ \right]\\\ \\ \Longrightarrow \ 1\ \times \frac{1}{cos\ ( 0)}\\\ \\ \Longrightarrow 1\

(05) Find the limit of:
\lim _{x\rightarrow 1} \ \left[ x^{3} -x^{2} +1\ \right] Put\ x=1\ in\ the\ given\ equation\\\ \\ \Longrightarrow \ 1^{3} -1^{2} +1\\\ \\ \Longrightarrow 1

Hence 1 is the limit of the given expression

## Proving Limit Formulas

(01) Prove the following expression step by step
\ \lim _{x\rightarrow a} \ \left[\frac{x^{n} -a^{n}}{x-a}\right] =na^{n-1} \\\ \\ Taking\ LHS\\\ \\ \Longrightarrow \lim {x\rightarrow a} \ \left[\frac{x^{n} -a^{n}}{x-a}\right] \ \ \ \ --eq( 1)\\\ \\ \\\ \\ Using\ Binomial\ Formula,\\\ \\ x^{n\ } -a^{n} =( x-a) \ ( x^{n-1} +ax^{n-2} +a^{2} x^{n-3} + \\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a^{3} x^{n-4} +\ .\ .\ .+a^{n-2} x+a^{n-1})\\\ \\ \\\ \\ Using\ the\ formula\ in\ eq( 1)\\\ \\ \Longrightarrow \lim_{x\rightarrow a} \ \left[\frac{( x-a) \ \left( x^{n-1} +ax^{n-2} .\ .\ .+a^{n-1}\right)}{x-a}\right] \\\ \\ \Longrightarrow \ \lim_{x\rightarrow a} \ \left[\left( x^{n-1} +ax^{n-2} .\ .\ .+a^{n-1}\right)\right] \\\ \\ Putting\ x=a\ for\ limit\ calculation,\ we\ get\ \\\ \\ \Longrightarrow \left( a^{n-1} +a^{n-1} +a^{n-1} .\ .\ .+a^{n-1}\right) \\\ \\ \Longrightarrow \ n\ a^{n-1}\\\ \\ L.H.S=R.H.S \\\ \\ Hence\ proved\

(2) Prove \ \lim _{x\rightarrow 0} \ \left[\frac{1-cos( x)}{x}\right] =0\\\ \\ Taking\ L.H.S \\\ \\ \Longrightarrow \lim_{x\rightarrow 0}\ \ \ \left[\frac{1-cos( x)}{x}\right]\\\ \\ \ \ We\ know\ that\\ \\ \ 1-cos( x) \ =2\ sin^{2}\left(\frac{x}{2}\right)\\\ \\ \\\ \\ \ Putting\ the\ value\ in\ eq( 1)\\\ \\ \ \Longrightarrow \lim_{x\rightarrow 0} \ \left[\frac{2\ sin^{2}\left(\frac{x}{2}\right)}{x}\right]\\\ \\ \Longrightarrow \lim_{x\rightarrow 0} \ \left[\frac{\ sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\right] \times sin\left(\frac{x}{2}\right)\\\ \\ \ \Longrightarrow \ \lim_{x\rightarrow 0} \ \left[\frac{\ sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\right] \times \lim_{x\rightarrow 0} sin\left(\frac{x}{2}\right) \ \ \ .\ .\ .eq( 2)\\\ \\ \\\ \\ \ \ \ we\ know\ that\ \lim_{x\rightarrow 0} \ \left[\frac{\ sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\right] \ =1\\\ \\ Applying\ the\ formula\ in\ eq( 2)\\\ \\ \Longrightarrow \ 1\ \times \lim _{x\rightarrow 0} sin\left(\frac{x}{2}\right) \\\ \\ putting\ x=0\ in\ the\ limit\\\ \\ \Longrightarrow \ \ 1\ \times sin\left(\frac{0}{2}\right)\ \Longrightarrow \ 1\ \times 0\ \\\ \\ \Longrightarrow \ 0\

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