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A function is said to be continuous if its graph can be drawn without lifting the pencil. So naturally, the discontinuity of a function occurs where we find breaks in the graph. Let us first see how continuity is defined in calculus.

A function f(x) is continuous at x =a, when

i. f(a) is defined

ii. $\lim_{x \to a} {f}(x)$ exists and

iii. $\lim_{x \to a} {f}(x)$ = f(a)

Indeed the condition iii determines the continuity of the function at a point x=a. Whenever (iii) is true, the other conditions should hold.

A function f(x) is discontinuous at x=a, when

i. f(a) is not defined

ii. $lim_{x \to a} {f}(x)$ does not exist

iii. $lim_{x \to a} {f}(x)$ $\neq$ f(a)

The type of discontinuity present is determined by the condition that fails.

**The three types of discontinuity seen in a function are **

**1. Point discontinuity or a hole**

**2. Infinite discontinuity**

**3. Jump discontinuity**

**4. Oscillating discontinuity**

Each of the above discontinuity are discussed with explanations supported by relevant graphs

i. f(a) is not defined

ii. $lim_{x \to a} {f}(x)$ does not exist

iii. $lim_{x \to a} {f}(x)$ $\neq$ f(a)

The type of discontinuity present is determined by the condition that fails.

Each of the above discontinuity are discussed with explanations supported by relevant graphs

A hole or point discontinuity is seen in the graph of a function when the first condition for continuity fails, that is f(a) is not defined or does not exist. Let us consider a few examples for point discontinuity.

Consider the piecewise defined function f(x) = $\begin{cases}

x^{2} & \text{ if } x<0 \\

2x & \text{ if } x>0

\end{cases}$

The graph to the left of x =0 is a parabola and the graph to the right is a straight line.

The function is not defined at x =0.

The left and right limits are computed as follows:

$\lim_{x \to 0^{-}}{f}(x)$ = $\lim_{x \to 0} x^{2}$ = 0

$\lim_{x \to 0^{+}}{f}(x)$ = $\lim_{x \to 0} 2x$ = 0Hence both the limits exist and hence $\lim_{x \to 0}{f}(x)$ and = 0

Hence the hole occurs due to non definition of a function value. It is a removable discontinuity.

The discontinuity can be removed by redefining the function in any one manner as shown below.

f(x) = $\begin{cases}

x^{2} & \text{ if } x<0 \\

2x & \text{ if } x>0

\end{cases}$

The existence of the limit makes it possible to define the function value equal to the limit and weakens the discontinuity.

Let us now look at a rational function where discontinuity occurs when the function is not defined.

f(x) = $\frac{x^{2}-1}{x-1}$ The graph of the function is given below and the discontinuity is explained

f(1) is not defined as x =1 makes the denominator zero. Note that x-1 is a factor of the numerator x^{3} -1.

X^{3} -1 = (x-1)(x^{2 }+x + 1)

For computing the left and right limits the reduced function x^{2} + x +1 can be used. Using methods to find left and left limits it can be show

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

Hence exists and $\lim_{x \to 1}{f}(x)$ = 3

Hence the point discontinuity occurs due to non definition of the function value. It is a removable discontinuity.

The discontinuity can be removed by redefining the function as

f(x) = x^{2} +x + 1

Thus the point discontinuity occurs due to the failure of the first condition for continuity that is when the function is not defined at the point, but the limit exists.

Consider the piecewise defined function f(x) = $\begin{cases}

x^{2} & \text{ if } x<0 \\

2x & \text{ if } x>0

\end{cases}$

The graph to the left of x =0 is a parabola and the graph to the right is a straight line.

The function is not defined at x =0.

The left and right limits are computed as follows:

$\lim_{x \to 0^{-}}{f}(x)$ = $\lim_{x \to 0} x^{2}$ = 0

$\lim_{x \to 0^{+}}{f}(x)$ = $\lim_{x \to 0} 2x$ = 0Hence both the limits exist and hence $\lim_{x \to 0}{f}(x)$ and = 0

Hence the hole occurs due to non definition of a function value. It is a removable discontinuity.

The discontinuity can be removed by redefining the function in any one manner as shown below.

f(x) = $\begin{cases}

x^{2} & \text{ if } x<0 \\

2x & \text{ if } x>0

\end{cases}$

The existence of the limit makes it possible to define the function value equal to the limit and weakens the discontinuity.

Let us now look at a rational function where discontinuity occurs when the function is not defined.

f(x) = $\frac{x^{2}-1}{x-1}$ The graph of the function is given below and the discontinuity is explained

f(1) is not defined as x =1 makes the denominator zero. Note that x-1 is a factor of the numerator x

X

For computing the left and right limits the reduced function x

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

Hence exists and $\lim_{x \to 1}{f}(x)$ = 3

Hence the point discontinuity occurs due to non definition of the function value. It is a removable discontinuity.

The discontinuity can be removed by redefining the function as

f(x) = x

Thus the point discontinuity occurs due to the failure of the first condition for continuity that is when the function is not defined at the point, but the limit exists.

The rational functions have infinite discontinuity at the point where they have a vertical asymptote.

The graph of the function f(x) = $\frac{x}{x-2}$ is given below and the infinite discontinuity, which is not removable is explained.

f(2) is not defined. The graph has a vertical asymptote at x =2. Note the function expression cannot be simplified, as x-2 is not a factor of the numerator.

The Left and right limits at x =2 can be computed as

$\lim_{x \to 2^{-}}{f}(x)$ = $-\infty$ and $\lim_{x \to 2^{+}}$ = $\infty$

$\lim_{x \to 2^{-}}{f}(x)$ $\neq$ $\lim_{x \to 2^{+}}$ and hence does not exist.The infinite discontinuity is the result of the first two conditions not holding true. The graph dips down to the left of the vertical asymptote x=2, and appears from top to its right. The discontinuity is not removable as the limit of the function does not exist at x =2.

Another example for infinite discontinuity explains the non existence of the limit in a different way.

The graph of the rational function f(x) = $\frac{1}{x^{2}}$ is seen here. f(0) is not defined. Both the right and limit are equal to $\infty$ . So it can be said that $\lim_{x \to 0}{f}(x)$ = $\infty$. Even though this still means that the limit does not exist, this is equivalent to saying that the function increases without bounds on either side of the vertical asymptote which makes the limit undetermined. This is also a non removable discontinuity as the limit does not exist at x =0.

The graph of the function f(x) = $\frac{x}{x-2}$ is given below and the infinite discontinuity, which is not removable is explained.

f(2) is not defined. The graph has a vertical asymptote at x =2. Note the function expression cannot be simplified, as x-2 is not a factor of the numerator.

The Left and right limits at x =2 can be computed as

$\lim_{x \to 2^{-}}{f}(x)$ = $-\infty$ and $\lim_{x \to 2^{+}}$ = $\infty$

$\lim_{x \to 2^{-}}{f}(x)$ $\neq$ $\lim_{x \to 2^{+}}$ and hence does not exist.The infinite discontinuity is the result of the first two conditions not holding true. The graph dips down to the left of the vertical asymptote x=2, and appears from top to its right. The discontinuity is not removable as the limit of the function does not exist at x =2.

Another example for infinite discontinuity explains the non existence of the limit in a different way.

The graph of the rational function f(x) = $\frac{1}{x^{2}}$ is seen here. f(0) is not defined. Both the right and limit are equal to $\infty$ . So it can be said that $\lim_{x \to 0}{f}(x)$ = $\infty$. Even though this still means that the limit does not exist, this is equivalent to saying that the function increases without bounds on either side of the vertical asymptote which makes the limit undetermined. This is also a non removable discontinuity as the limit does not exist at x =0.

Infinite discontinuity is caused due to the failure of both the first and second conditions (and therefore the third). Let us look at the instances where the function is defined, but the limit does not exist.

Consider the piecewise function f(x) =$\begin{cases}

x^{2}-1& \text{ if } x< 0 \\

2x& \text{ if } x\geq 0

\end{cases}$

The graph of f(x) is shown here with the jump visible.

The graph of the function is the parabola y =x^{2} -1 for x < 0 and the line through the origin for x $\neq$ 0.

f(0) = 0 by virtue of y =2x.

The left limit on the parabola side is

$\lim_{x \to 0^{-}}{f}(x)$ = -1

The right limit taken along the line is

$\lim_{x \to 0^{+}}{f}(x)$=0

Since the two limits are not equal

$\lim_{x \to 0}{f}(x)$ does not exist.As the parabola approaches the vertex (0-1), the graph jumps to (0,0) and continues along the line y =2x. This is again an instance of non removable discontinuity.

Jump discontinuity dominates the graphs of the greatest integer and the smallest integer functions.

The following graph of the function f(x) = Sin $(\frac{\pi}{x})$.

The function oscillates rapidly near x=0. Both the function value and the limit value cannot be determined. Oscillating discontinuity is again a non removable discontinuity, though it is not visible in the graph clearly as other cases of discontinuities.

From the study done here, we can conclude a discontinuity becomes removable, if the limit at the point of discontinuity exists. Point discontinuity (Hole) can be removed by defining the function value equal to the limit, as the limit exists in this case.

Consider the piecewise function f(x) =$\begin{cases}

x^{2}-1& \text{ if } x< 0 \\

2x& \text{ if } x\geq 0

\end{cases}$

The graph of f(x) is shown here with the jump visible.

The graph of the function is the parabola y =x

f(0) = 0 by virtue of y =2x.

The left limit on the parabola side is

$\lim_{x \to 0^{-}}{f}(x)$ = -1

The right limit taken along the line is

$\lim_{x \to 0^{+}}{f}(x)$=0

Since the two limits are not equal

$\lim_{x \to 0}{f}(x)$ does not exist.As the parabola approaches the vertex (0-1), the graph jumps to (0,0) and continues along the line y =2x. This is again an instance of non removable discontinuity.

Jump discontinuity dominates the graphs of the greatest integer and the smallest integer functions.

The following graph of the function f(x) = Sin $(\frac{\pi}{x})$.

The function oscillates rapidly near x=0. Both the function value and the limit value cannot be determined. Oscillating discontinuity is again a non removable discontinuity, though it is not visible in the graph clearly as other cases of discontinuities.

From the study done here, we can conclude a discontinuity becomes removable, if the limit at the point of discontinuity exists. Point discontinuity (Hole) can be removed by defining the function value equal to the limit, as the limit exists in this case.