Continuity of Functions

In mathematics, continuity describes how smoothly a function behaves without sudden jumps, breaks, or holes. It is an important unit of Calculus as it forms the base, and it helps us further to prove whether a function is differentiable or not.

Formally, a function f(x) is continuous at a point x = a if three conditions hold:

• f(a) is defined (the function has a value at a).
• lim⁡_x→a f(x) exists (the limit from both sides exists).
• lim⁡_x→a f(x) = f(a) (the limit equals the actual value).

If these conditions are satisfied at every point in an interval, the function is continuous on that interval. If they hold for all real numbers in its domain, the function is said to be continuous everywhere.

The above three conditions are respectively called Function Condition, Limit Condition, and Point Condition.

Note: Discontinuity occurs when any of the above conditions fail, leading to breaks, holes, or jumps in the curve

The best example of Continuous functions is Trigonometric functions such as sin(x) and cos(x). They are periodic functions and the values of the functions exist at each point.

One-Sided Continuity

One-Sided Continuity is the continuity in which the input approaches the point either from the left side or right side respectively to the value of the function at that input. We usually use + or - as an exponent on the input. 

For the left side, we use the negative sign and this left side is known as Left Hand Limit abbreviated as LHL. The LHL is denoted by Iim_x→a-f(x). 

For the right side, we use the positive sign. The right side is also known as Right Hand Limit. It is denoted by Iim_x→a+f(x)

In simpler terms, continuity means that there are no sudden jumps, holes, or breaks in the graph of the function. The function can be traced without lifting the pen from the paper.

Continuity of Function Examples

Let us check the continuity of the function f(x) = sin (x) at a = 0

Let us find the value of the function at a=0

f(0) = sin(0) = 0

Now calculating the Left-Hand limits and the Right Hand Limits respectively we get

LHL = lim_{x → 0-} f(x)
⇒ LHL = lim_{h → 0} f(0 - h)
⇒ LHL = lim_{h → 0} sin(-h)
⇒ LHL = - lim_{h → 0} sin(h)
⇒ LHL = 0

RHL = lim_{x → 0+} f(x)
⇒ RHL = lim_{h → 0} f(0 + h)
⇒ RHL = lim_{h → 0} sin(h)
⇒ RHL = lim_{h → 0} sin(h)
⇒ RHL = 0

The condition Iim_(x→0-)f(x)=f(0)=Iim_(x→0+)f(x) is satisfied. Hence the function is continuous at x = 0

Note: All Polynomial, logarithmic, and exponential functions such as ex are continuous in all the domains.

Techniques for Determining Continuity

There are different techniques for determining Continuity. Some of them are as follows:

Algebraic Manipulation

Here we use Algebraic tricks such as factorization, the use of Trigonometric Identities, etc to solve continuity-related problems. Let us elaborate with the help of an example.

Example: Let f(x) be defined as follows

\bold{f(x) = \begin{cases}\frac{x^2 - 5}{x - \sqrt5} & \text{if} ~x ≠ \sqrt5\\ 2√5 & \text{if} ~x = \sqrt5\end{cases}}

Test the continuity at x = √5.

Solution:

If we put x = √5 in the function (x² - 5) / (x - 5) then the function is undefined. 

Therefore we need to factorize and simplify the function.

lim_x→√5 (x² - 5) / (x - √5)

= lim_x→√5 [(x - √5)(x + √5)]/(x - √5)

= lim_x→√5 (x + √5) = 2√5

And, as at x = √5, function is given by f(x) = f(√5)

Hence the function is continuous.

Piecewise Function

A piecewise function is a function that is defined differently for different functions and is said to be continuous if the graph of the function is continuous at some intervals. Let's consider an example to understand it better.

Example: Let f(x) be defined as follows.

\bold{f(x) = \begin{cases} x + 2, & \text{if } x < 1 \\ ~~~~1, & \text{if } x = 1 \\ 2 - x, & \text{if } x > 1\end{cases}}

Test the continuity at x = 1.

Solution:

Let us calculate the Left Hand Limit

LHL = lim_{x → 1-} f(x) 
⇒ LHL = lim_{h → 0}  f(1 - h)
⇒ LHL = lim_{h → 0} (1 - h + 2)
⇒ LHL = lim_{h → 0}  (3 - h)
⇒ LHL = 3

f(1) = 1 is given

and,

RHL = lim_{x → 1+} f(x)
⇒ RHL = lim_{h → 0} f(1 + h)
⇒ RHL = lim_{h → 0} (2 - (1 + h))
⇒ RHL = lim_{h → 0} 1
⇒ RHL = 1

The Right Hand Limit is equal to f(1) but is not equal to Left Hand Limit. Hence the function is not continuous.

Rational Functions

Rational functions are of the form p/q where q is not equal to 0. If radicals are present we can check the continuity of these types of functions by multiplying the numerator and denominator with the conjugate of the denominator.

The example is as follows:

Example: Let f(x) be defined as follows.

\bold{f(x) = \begin{cases}\frac{\sqrt{x+1}-1}{x } & \text{if} ~x ≠ 1\\ 0.5 & \text{if} ~x = 1\end{cases}}

Check the continuity at x = 1.

Solution:

lim_{x → 0} (√(x + 1) - 1) / x
=lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} \times \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}\\ =lim_{x\to 0}\frac{x+1-1}{x}\times\frac{1}{(\sqrt{x+1}+1} \\ =lim_{x\to 0} \frac{1}{(\sqrt{x+1}+1}\\
= 1/2 = 0.5

And f(1) = 0.5

Hence it is continuous at x=1.

Discontinuity

If any of the continuity conditions fails, then the function is said to be discontinuous. Alternatively, it can be said that if there occurs a break when a part is drawn on the graph paper in a given interval, then also it defines discontinuity. There are different kinds of Discontinuity.

Jump Discontinuity

This is the discontinuity in which the limit values of the function are different. It means that the Left Hand Limit and the Right Hand Limit exist but the values are different thereby not satisfying the condition Iim_x→a-f(x) = Iim_x→a+f(x)

Let us illustrate with the help of an example

For a function f(x), if 

lim_{x → 3-} f(x) = 3, and

lim_{x → 3+} f(x) = 8

We can see the Left Hand Limit and the Right Hand Limit exists but they are not the same. 

Hence this is a jump discontinuity.

Point Discontinuity

This discontinuity occurs in rational expressions in which both the numerator and the denominator become 0. However, this discontinuity can be removed. Let us illustrate with the help of an example

Example: Check the continuity of the following function f(x) at x = 1,

\bold{f(x) = \frac{x^2 - 1}{x - 1}}

Solution:

This is a point discontinuity since if we put x =1 in the function it is of the form 0/0. However, by factorizing this discontinuity can be removed.

LHL = lim_{x → 1-} f(x)
⇒ LHL = lim_{x → 1-} [(x - 1)(x + 1) / (x - 1)]
⇒ LHL = lim_{x → 1-} (x + 1)
⇒ LHL = lim_{h → 0} (1 - h + 1)
⇒ LHL = 2

and 

RHL = lim_{x → 1+} f(x)
⇒ RHL = lim_{x → 1+} [(x - 1)(x + 1) / (x - 1)]
⇒ RHL = lim_{x → 1+} (x + 1)
⇒ RHL = lim_{h → 0} (1 + h + 1)
⇒ RHL = 2

Hence, it is continuous at x = 1.

Infinite Discontinuity

This discontinuity occurs when either the value of the left-hand limit tends to negative infinity or the value of the right-hand limit tends to infinity. Let us illustrate with the help of an example.

Example: Check the continuity of the following function at x = 0,

\bold{f(x)=\frac{1}{x}}

Solution:

LHL = lim_{x → 0-} f(x)
⇒ LHL = lim_{x → 0-} (1 / x)
⇒ LHL = lim_{h → 0} (1 / (0 - h))
⇒ LHL = -∞

and, 

RHL = lim_{x → 0+} f(x)
⇒ RHL = lim_{x → 0+} (1 / x)
⇒ RHL = lim_{h → 0} (1 / (0 + h))
⇒ RHL = +∞

Thus, at x = 0 function has Infinite Discontinuity.

Solved Examples on Continuity

Example 1: For function f(x) defined as 

\bold{f(x) = \begin{cases} 4x, & \text{if } x < 2 \\ ~~~~8, & \text{if } x = 2 \\ 3x+2, & \text{if } x > 2\end{cases}}

Check the continuity at x =2.

Solution:

We will calculate the limits at x = 2

LHL = Iim_(x→2-) f(x) = Iim_(h→0) f(2-h)

⇒ LHL = Iim_(h→0)4×(2-h)

⇒ LHL = 4×(2-0)

⇒ LHL = 8

RHL = Iim_(x→0+) f(x) = Iim_{(h → 0)} f(0+h)

⇒ RHL = Iim_{(h → 0)}3×(2+h)+2

⇒ RHL = 3×(2+0)+2

⇒ RHL = 8

Example 2: Find the value of 'm' at which the function is continuous at x = 9

\bold{f(x) = \begin{cases}mx+5 & \text{if} ~x ≠ 9\\ 8x & \text{if} ~x = 9\end{cases}}

Solution:

lim_{x → 9-} f(x) = lim (h → 0) f(9 - h)
LHL = lim_{h → 0} [m × (9 - h) + 5]
LHL = 9m + 5

RHL = lim_{x → 9+} f(x)
⇒ RHL = lim_{h → 0} f(9 + h)
⇒ RHL = lim_{h → 0} [8 × (9 + h)]
⇒ RHL = 72

For a function to be continuous Left Hand Limit is equal to Right Hand Limit

Thus, 9m + 5 =72

⇒ m = 7.4444

For the function to be continuous the value of m should be 7.44.

Example 3: Test the continuity at x = 1, for the following function.

\bold{f(x) = \begin{cases}\frac{x^2-3x+2}{x-1} & \text{if} ~x ≠ 1\\ 9x & \text{if} ~x = 1\end{cases}}

Solution:

To test the continuity of the function f(x) at x = 1, the following conditions must be checked:

1. f(1) is defined.
2. lim⁡_x→1 f(x) exists.
3. lim⁡_x→1 f(x) = f(1).

Step 1. Evaluate f(1):

Given that f(x) = 9x  when x = 1:
f(1) = 9⋅1 = 9
Thus, f(1) is defined and equals 9.

Step 2. Evaluate lim⁡_x→1 f(x)

For x ≠ 1, f(x) = \frac{x^2 - 3x + 2}{x - 1}
x = 1 gives undefined value (i.e., 0/0)

Factorize and simplify the function.
x² − 3x + 2 = (x − 1)(x − 2)

Thus, for x ≠ 1,
f(x)= \frac{(x−1)(x−2)}{x -1 }= x -2

Therefore, lim⁡_x→1 f(x) = lim⁡_x→1 (x − 2) = 1 − 2 = −1
The limit exists and equals −1.

Step 3. Compare lim⁡_x→1 f(x) and f(1)

• lim _x→1 ​f(x) = −1
• f(1) = 9
  Since −1 ≠ 9, the limit does not equal the function value at x = 1.

The function f(x) is discontinuous at x = 1 because lim_⁡x→1 f(x) ≠ f(1).

Example 4: \bold{f(x) = \begin{cases}\frac{sin(x-2)}{x^2-4} & \text{if} ~x ≠ 2\\ 0.25 & \text{if} ~x = 2\end{cases}}

Test the continuity at x = 2 of the above-mentioned function.

Solution:

If we put x = 2 in the function sin(x - 2) / (x² - 4) then the function is undefined. Therefore we need to factorize and simplify the function.

\lim_{x\to 2}f(x) = \lim_{x\to 2} \frac{sin(x - 2)}{x^2 - 4} \\ \Rightarrow \lim_{x\to 2} f(x) = \lim_{x\to 2} \frac{sin(x - 2) }{(x - 2)\times (x + 2)} \\ \Rightarrow \lim_{x\to 2} f(x) = \lim(x\to 2) \frac{1}{x + 2}\times \lim(x\to 2) \frac{sin(x - 2)}{x - 2} 

⇒ lim_x→2 ​f(x) = 1/4 = 0.25
⇒ lim_x→2 ​f(x) = f(2) = 0.25

Thus, the function is continuous at x = 2.

Example 5: Find the relation between m and n if the function is continuous at point x = -5

\bold{f(x) = \begin{cases}mx^2 & \text{if} ~x < -5\\ nx+5 & \text{if} ~x \geqslant -5\end{cases}}

Solution:

We will compute the Left Hand Limit and Right Hand Limit separately

LHL = lim_{x→−5 −} mx²
⇒ LHL = lim_h→0 m×(−5−h)²
⇒ LHL = 25m

RHL = lim _{x→−5 +} nx+5 
⇒ RHL = lim_h→0  n×(-5+h)+5 
⇒ RHL = -5n + 5

For a function to be continuous LHL = RHL

25m = -5n + 5 
⇒ 25m + 5n = 5
⇒ 5m + n = 1

Related Articles

• Continuity and Discontinuity
• Calculus in Maths
• Introduction to Limit
• Limits in Calculus - Definition, Formula, Examples
• Properties of Limits
• Real Life Applications of Discontinuity

Continuity of Functions: Definition, Types, Condition, Examples

Continuity of Functions: Definition, Types, Condition, Examples

Continuity (new)