Continuity and Discontinuity in Calculus

Continuity and discontinuity are fundamental concepts in calculus and mathematical analysis that describe the behavior of functions. A function is continuous at a point if you can draw the graph of the function at that point without lifting your pen from the paper. A function is discontinuous at a point x = c if it fails to be continuous at that point.

Continuity

A function is said to be continuous if one can sketch its curve on a graph without lifting the pen even once.

A function is said to be continuous at x = a if and only if the three following conditions are satisfied:

The function is defined at x = a.

The limit of the function as x approaches a exists.

The limit of the function as x approaches a is equal to the function value at x = a.

Continuity on an Open Interval (a,b)

A function f(x) is said to be continuous on the open interval (a,b) if it is continuous at every point in the interval.

That is, for every c∈(a,b):\lim_{x \to c} f(x) = f(c)

Continuity on a Closed Interval [a,b]

A function f(x) is said to be continuous on the closed interval [a,b] if:

• It is continuous at every point in the open interval (a,b), and
• It satisfies one-sided continuity at the endpoints: At x=a: lim_x⇢a f(x) = f(a) and x=b: lim_x⇢b f(x) = f(b)

Example 1: Prove that the function f(x) = 5x - 3 is continuous at x = 0.

Given, f(x) = 5x - 3

At x = 0 , f(0) = (5 × 0) - 3 = -3 
lim_x⇢0 f(x) = lim_x⇢0 (5x - 3) = (5 × 0) - 3 = -3 
lim_x⇢0 f(x) = f(0) 
Therefore, f(x) is continuous at x = 0.   

Example 2: Examine the function f(x) = |x - 5|, for continuity.

Given function, f(x) = |x - 5| 
Domain of f(x) is real and infinite for all real x 
Here , f(x) = |x - 5| is a modulus function 
As , every modulus function is continuous 
Therefore , f(x) is continuous in its domain R.   

Example 3: Is the function f(x) =  x - sinx + 5  is continuous at x = π.

Given function is f(x) = x - sinx + 5 
L.H.L = lim_x⇢π- (x - sinx + 5) = lim_x⇢π- [(π - h) - sin(π - h) + 5] = π + 5 
R.H.L = lim_x⇢π+ (x - sinx + 5) = lim_x⇢π+ [(π + h) - sin(π + h) + 5] = π + 5 
And, f(π) = π - sinπ + 5 = π  + 5 

Since , L.H.L = R.H.L = f(π) 
Therefore , f(x) is continuous at x = π 

Example 4: Examine the continuity of the function f(x) = 2x - 1 at x = 3.

Given f(x) = 2x - 1 
At x = 3, f(x) = (2 × 3) - 1 = 5 
lim_x⇢3 f(x) = lim_x⇢3 f(x) = (2×3) - 1 = 5 
lim_x⇢3 f(x) = f(3) 
Therefore, f(x) is continuous at x = 3

Example 5: Examine the function is continuous or not?

For x > 0, y = x and x < 0, y = -x

So, We Know it is continuous for x > 0 and x < 0. To check if it is continuous at x = 0 , check the limit:

lim_x⇢0-  |x| = lim_x⇢0- (-x) = 0

lim_x⇢0+  |x| = lim_x⇢0+ (x) = 0

So, lim_x⇢0 |x| = 0 , which is equal to the value of the function at 0. Therefore, It is continuous everywhere. 

Discontinuity

A function is discontinuous at a point x = a if the function is not continuous at a. The function "f" is said to be discontinuous at x = a in any of the following cases:

1. f(a) is not defined
2. lim_x⇢a+ f(x) and lim_x⇢a-  f(x) exists, but are not equal.
3. lim_x⇢a+ f(x) and lim_x⇢a- f(x) exists and are equal but not equal to f(a).

Types of Discontinuity

There are three basic types of discontinuities

1. Removable(point) Discontinuity: The graph has a hole at a single x-value. Imagine you're walking down the road, and someone has removed a manhole cover. This is a category of discontinuity in which the function has a well defined two-sided limit at x = a, but either f(a) is not defined or f(a) is not equal to its limit.

• lim_x⇢a f(x) ≠ f(a)
• f(a) = lim_x⇢a f(x)

2. Infinite Discontinuity: The function goes toward positive or negative infinity. Imagine a road getting closer and closer to a river with no bridge to the other side. The function diverges at x = a to give it a discontinuous nature here. That is to say, f(a) is not defined. Since the value of the function at x = a tends to infinity or doesn’t approach a particular finite value, the limits of the function as x → a are also not defined.

3. Jump Discontinuity: The graph jumps from one place to another. Imagine a superhero going for a walk, he reaches a dead end and, because he can, flies to another road. In this type of discontinuity, the right-hand limit and the left-hand limit for the function at x = a exists; but the two are not equal to each other.

• lim_x⇢a+  f(x) ≠ lim_x⇢a- f(x)

Solved Examples

Example 1: Find all the points of discontinuity of the function f defined by f(x) = |x| - |x+1|.

Given function = |x| - |x+1|

From the function the critical points are x=0 and x=-1

For x< -1 , f(x) = -x - (-x-1) = 1

For -1<=x<0 , f(x) = -x-(x+1) = -2x-1

For x>= 0 , f(x) = x - (x+1) = -1

Checking the Discontinuity :

At x = -1

Left limit: lim_x⇢-1-f(x) = 1

Right limit: lim_x⇢-1+ f(x) = -2(-1)-1 = 1

Function value: f(-1) = 1

Continuous at x = -1 (limits and function value are equal )

At x = 0

Left limit: lim_x⇢-0-f(x) = -2(0) -1 = -1

Right limit: lim_x⇢-0+ f(x) = -1

Function value: f(0) = -1

Continuous at x = 0 (limits and function value are equal )

Since the function is continuous at x = -1 and x = 0, there are no points of discontinuity. Thus, f(x) is continuous everywhere.

Example 2: Show that the function defined by g (x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x

For x approaching an integer n from the left (x→ n^- ) : lim _x⇢n- g(x) = 1

because [x] = n - 1

For x approaching n from the right (x→n⁺): lim _x⇢n+ g(x) = 0

because [x] = n

Since the left-hand limit and the right-hand limit are not equal at any integer n, g(x) is discontinuous at all integers.