A function is continuous, if it has a break. Also, we can say that function of the graph which is not connected with each other is known as discontinuous function. The following is the example graph which represents the discontinuity of a function.

In the above graph, the limits of the function to the left and to the right are not equal and hence the limit at x = 3 does not exist anymore. Such function is said to be discontinuity of a function.

A function f(x) is said to be discontinuous at a point 'a' of its domain D if it is not continuous there. The point 'a' is then called a point of discontinuity of the function. The discontinuity may arise due to any of the following situation.

- The right hand limit or the left hand limit or both of a function may not exist.
- The right hand limit and the left hand limit of function may exist but are unequal.
- The right hand limit as well as the left hand limit of a function may exist, but either of the two or both may not be equal to f(a).

A function f(x) is said to have a discontinuity of the first kind at x = a if the left hand limit of f(x) and right hand limit of f(x) both exist but are not equal. f(x) is said to have a discontinuity of the first kind from the left at x = a if the left hand of the function exists but not equal to f(a). There are three types of discontinuity.

- Removable Discontinuity
- Jump Discontinuity
- Infinite Discontinuity

There are two types of jump discontinuity as follows:

f(x) is said to have a discontinuity of the first kind from the right at x = a, if the right hand of the function exists but not equal to f(a) and in Jump Discontinuity, the Left Hand Limit and the Right Hand Limit exist and are finite but not equal to each other.

A function f(x) is said to have discontinuity of the second kind at x = a, if neither left hand limit of f(x) at x = a nor right hand limit of f(x) at x = a exists.

Given below are some of the examples on Jump Discontinuity.

$\lim_{x \to 2}$ $\frac{|x - 2|}{x - 2}$

$\lim_{x \to 2}$ $\frac{+(x - 2)}{x - 2}$ = 1 and $\lim_{x \to 2}$ $\frac{-(x - 2)}{x - 2}$ = -1

If $\lim_{x \to 3}\left | x - 3 \right |\div (x - 3)$, find the left hand and right hand limits of the function at x = 3.

Let f(x) = $\left | x - 3 \right |\div (x - 3)$

$\lim_{x \to 3}\left | x - 3 \right |\div (x - 3)$

Here, two cases will exist that is ,

|x - 3| = +(x - 3) or |x - 3| = - (x - 3) x -> 3

Now, $\lim_{x \to 3^{+}}(x - 3)\div (x - 3) = 1$

Right hand limit is equal to 1

$\lim_{x \to 3^{+}} -(x - 3)\div (x - 3) = -1$

Left hand limit is equal to - 1

Let us see some of the examples on discontinuous functions:

In the graph, we can note that there is a discontinuity at the point 2, which gives a very big change in the given function.

Consider the given function: f(x) = $\frac{2}{(x^{2} - x)}$. Is this function discontinuous? If so, find at what point it is discontinuous.

Factorizing the given function, we obtain the result as

F(x) = $\frac{2}{(x^{2} - x)}$ = $\frac{2}{x(x-1)}$.

From this, it is clearly observe that the function is said to be discontinuous at x = 0 and x = 1.

Consider the rational function f(x) = $\frac{[(x+2) (x+3)]}{(x+3)}$. Is this function discontinuous? If so, find at what point it is discontinuous.

On simplifying the above function, we can get f(x) =x + 2. It looks like a simple linear function. Consider the original function without any simplification, and then we can notice the rational function.

At x = -3, the function becomes discontinuous.

Consider the following function:

y = x + 6, x < 0

y = $\frac{3}{(x^{2} + 1)}$, x $\geq$ 0

This function has two different equations. First equation became discontinuous at x = -6 and second equation became discontinuous at x = -1.

A function f(x) is said to have removable discontinuity at x = a, if left hand limit at x tends to a is equal to right hand limit at x tends to a but their common value is not equal to f(a)

A removable discontinuity occurs when there is a rational expression with common factors in the numerator and denominator. Since, these factors can be canceled the discontinuity is removed.

Given below are some of the examples on removable discontinuity.

$\lim_{x \to 0} f(x) = \lim_{x \to 0} $$[\frac{\sin 2x}{x}]$

= $\lim_{2x \to 0}$$[\frac{\sin 2x}{2x}]$$ \times 2$ [Multiplying and dividing with 2]

= $1 \times 2$

= $2$

= $\lim_{2x \to 0}$$[\frac{\sin 2x}{2x}]$$ \times 2$ [Multiplying and dividing with 2]

= $1 \times 2$

= $2$

Given function is $f(x) = (x^{2} - 8x + 15) \div (x^{2} - 6x + 5)$

Here, if $x = 5$ is plugged in for $x$ of $f(x)$, then both the numerator and the denominator values becomes 0.

That is, we get (0/0) at x = 5. That means at x = 5 the limit of the function does not exist.

But, if the functions are factorized and simplified by eliminating the common factor, the discontinuity is removed. Hence, the limit of the function can be obtained.

Numerator = x^{2} - 8x + 15

= x^{2} - 5x - 3x + 15

= x(x - 5) - 3(x - 5)

= (x - 5) (x - 3)

And, Denominator = x^{2} - 6x + 5

= x^{2} - 5x - x + 5

= x(x - 5) - 1(x - 5)

= (x - 5) (x - 1)

Here, we see that (x - 5) is the common factor and this can be canceled.

So, after simplifying the given function $f(x)$ = $\frac{(x - 3)}{(x - 1)}$

$\lim f(x)$ at $x = 5$, is $\frac{5 - 3}{5 - 1} = \frac{2}{4} =\frac{1}{2}$

Here, if $x = 5$ is plugged in for $x$ of $f(x)$, then both the numerator and the denominator values becomes 0.

That is, we get (0/0) at x = 5. That means at x = 5 the limit of the function does not exist.

But, if the functions are factorized and simplified by eliminating the common factor, the discontinuity is removed. Hence, the limit of the function can be obtained.

Numerator = x

= x

= x(x - 5) - 3(x - 5)

= (x - 5) (x - 3)

And, Denominator = x

= x

= x(x - 5) - 1(x - 5)

= (x - 5) (x - 1)

Here, we see that (x - 5) is the common factor and this can be canceled.

So, after simplifying the given function $f(x)$ = $\frac{(x - 3)}{(x - 1)}$

$\lim f(x)$ at $x = 5$, is $\frac{5 - 3}{5 - 1} = \frac{2}{4} =\frac{1}{2}$

Given function is $f(x)$ = $\frac{x^{2} - 4}{x - 2}$

Here, if x = 2 is plugged in for x of f(x) then both the numerator and the denominator values becomes 0.

That is, we get (0/0) at x = 2. That means, at x = 2 the limit of the function does not exist. But, if the functions are factorized and simplified by eliminating the common factor, the discontinuity is removed. Hence, the limit of the function can be obtained.

Numerator = x^{2} - 4

= x^{2} - 2^{2}

= (x - 2) (x + 2)

And, Denominator = x - 2

Here, we see that (x - 2) is the common factor and this can be canceled.

So, after simplifying the given function f(x) = (x + 2)

Limit of f(x) at x = 2, is (2 + 2) = 4

Here, if x = 2 is plugged in for x of f(x) then both the numerator and the denominator values becomes 0.

That is, we get (0/0) at x = 2. That means, at x = 2 the limit of the function does not exist. But, if the functions are factorized and simplified by eliminating the common factor, the discontinuity is removed. Hence, the limit of the function can be obtained.

Numerator = x

= x

= (x - 2) (x + 2)

And, Denominator = x - 2

Here, we see that (x - 2) is the common factor and this can be canceled.

So, after simplifying the given function f(x) = (x + 2)

Limit of f(x) at x = 2, is (2 + 2) = 4

Here,
if 8 is substituted for x, then the denominator becomes zero and hence
the function will not exist. But, if the functions factorized to its
simplest form before substitution, the Discontinuity can be
removed.

$\lim_{x \to 8}$ $\frac{(x + 8)(x - 8)}{(x - 8)}$

$\lim_{x \to 8} (x + 8)$

= 8 + 8

= 16

Its graph is as follows:

$\lim_{x \to a} $$\frac{(x + 3)}{(x + 3)^{2}}$ By plugging-in $x = -3$

= $\frac{(-3 + 1)}{(-3 + 3)^{2}}$

= $\infty$The function is discontinuous because it is possible for the denominator to equal zero at x = 1. This means that f(1) does not exist, and the function has an asymptote at x = 1.

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Jump Discontinuity | Dimensional Space |

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