A function that doesn’t have any breaks or jumps in it is called continuous.
You can liken the curve of a function to a bridge spanning across a river. If you can drive across from the beginning to the end of the bridge without any interruptions, then the bridge is “continuous”.
Let’s say that at some location on the bridge, the concrete withers and falls into the river. Now there is a big gap in the middle of the bridge. Can you still drive your car over the gap to reach the other side? Very unlikely. The bridge in this case is not ‘continuous’. Think of this scenario as a removable discontinuity, or hole, of a function.
Let’s say that the construction workers didn’t build the bridge correctly and the two ends of the bridge do not connect. Would you still be able to drive across the bridge to the other side? Again, very unlikely. The bridge is not ‘continuous’. Think of this scenario as a jump discontinuity of a function.
Let’s say that the bridge opens up to make way for passing boats. Can you still drive your car across to reach the other side? Very unlikely. If anything, you would drive up into the air to nowhere (until you reach the edge, that is). The bridge is not ‘continuous’. Think of this scenario as an infinite discontinuity of a function.
Now we will discuss how any of the functions can be continuous at a specific point.
Definition: Continuity
A given function f(x) is continuous at the value x = c under three conditions:

To summarize, for a function to be continuous at an x = c, there cannot be any discontinuities of any type (removable, jump, or infinite), there needs to be a limit L at x = c, and there needs to be a defined y value at x = c.
Example 1: Determine if the given function is continuous or not at x = 3, x = 4 and x = 2.
Solution:
First, we will check the continuity of a function at x = 4.
From the graph, we will see that the lim _{x→4} F(x) = 3. We also see that F(4) = 3. Therefore, the graph is continuous at x = 4.
Next, let’s check the continuity at x = 3.
From the graph, the lim_{ x→3} F(x) = 0. However, F(3) does not exist because there is a removable discontinuity (hole). Therefore, the graph is not continuous at x = 3.
Lastly, let’s check the continuity at x = 2.
From the graph, we see that F(2) = 1. However, the lim _{x→2} F(x) does not exist because the limit from both the left and right side are different (y = 1 and y = 1, respectively). Therefore, the graph is not continuous at x = 2.
Piecewise Functions and Continuity
Piecewise functions are functions defined by 2 or more subfunctions. Each subfunction has a subinterval (or subdomain), which makes up the entire interval of the function.
An example of a piecewise function is below:
A piecewise function is continuous on a specified interval if the following conditions are achieved:
 It is defined throughout that interval.
 The individual functions are continuous on their subintervals.
 There is no discontinuity at each endpoint of the subintervals.
Example 2
Our piecewise function F(x) is defined as follows:
Determine whether the function is continuous over the entire interval. If not, determine the x value(s) where the function is not continuous.
Solution
We need to find the limits of each region from left to right:
First, let’s determine the limit of 2x as x approaches zero from the left side (from ∞ to 0):
(1)^{2}+1
= 1 + 1
= 2
Fourth, let’s determine the limit of 7 – x as x approaches 1 from the right side (from ∞ to 1):
Since the limit from both the left and right side of the function as x approaches 1 are different, there is no continuity at x = 1.
Fifth, we need to determine the continuity of our last subfunction, 7 – x, on the interval (1, ∞). Since 7 – x is a linear function, we know the domain includes all real numbers. Therefore, the subfunction will not experience any discontinuities as it extends to ∞.
Overall, the function would be continuous over the entire interval EXCEPT at x = 0 and x = 1.
Below is a graph that shows the discontinuity:
Example 3
Our piecewise function F(x) is defined as follows:
Determine whether the function is continuous over the entire interval. If not, determine the x value(s) where the function is not continuous.
Solution
We need to find the limits of each region from left to right:
First let’s determine the limit of x + 6 as x approaches zero from the left side:
Since the limit from both the left and right side of the function as x approaches zero are the same, the function will be continuous at x = 0.
Third, let’s determine the limit of 6as x approaches 2 from the left side:
Since the limit from both the left and right side of the function as x approaches 2 are different, there is no continuity at x = 2.
Fifth, we need to determine the continuity of our last subfunction, , on the interval (2, ∞) Since is a rational function, we have to set the denominator equal to zero to see which x values make the function discontinuous.
First, let’s factor the denominator:
x^{2}+2x3= (x1)(x+3)
Set each factor equal to zero:
x – 1 = 0
x + 3 = 0
Solve each equation:
x = 1
x = 3
Therefore, the function will not be continuous on x = 1 and x = 3. However, on the interval of (2, ∞), these discontinuities would not apply. Therefore, on the interval of (2, ∞), the function would be continuous.
To summarize, the function will be continuous over the entire interval EXCEPT at x = 2.
Below is a graph that shows the discontinuity:
Example 4
Our piecewise function F(x) is defined as follows:
Determine the value of c that will make the function continuous.
Solution
First, let’s set the subfunctions equal to each other:
cx + 2 = 4x + c
Because both functions need to be continuous at x = 2, substitute x = 2 inside the equation:
c(2) + 2 = 4(2) + c
Solve for c:
2c + 2 = 8 + c
Subtract c from both sides,
2c – 1c + 2 = 8
1c + 2 = 8
Subtract 2 on both sides,
1c + 2 – 2 = 8 – 2
1c = 6
c = 6
6 is the value of c that will make both functions continuous at x = 2.
To check our work, let’s substitute c = 6 and x = 2 into both subfunctions and see if they are equal to each other:
cx + 2 = 4x + c
(6)(2) + 2 = 4(2) + (6)
12 + 2 = 8 + 6
14 = 14
Our calculations were correct.
Let’s graph the piecewise function to further confirm our results:
Determining Intervals of Continuity of Functions
Example 5: Determine the intervals of continuity of the following functions:
Let’s start with f(x) = 2x^{2}3x. Since f(x) is a polynomial, the function will be continuous over its entire domain (∞, ∞).
Let’s examine g(x) =. Since g(x) is a rational function, we have to set the denominator equal to zero in order to see where the function is not continuous:
x – 4 = 0
x = 4 (vertical asymptote)
The function would be continuous across its domain EXCEPT at x = 4. The intervals of continuity are (,∞ 4) (4, ∞).
Let’s examine h(x) =. Because h(x) is a rational function, we need to set the denominator equal to zero to see where the function is not continuous.
x^{2}4x+3=0
Factor the denominator:
(x3)(x1) = 0
Set each factor equal to zero:
x – 3 = 0
x – 1 = 0
Solve each equation:
x = 3 (vertical asymptote)
x = 1 (vertical asymptote)
The function would not be continuous at x = 1 and x = 3. Therefore, the intervals of continuity of h(x) are
Summary of Section
Discontinuity: A region in the graph where f(x) experiences a distinct break or interruption. There are 3 common types of discontinuities:
Continuity: A given function f(x) is continuous at the value x = c under three conditions:
F(c) = L. 2. The limit of F(x) as x approaches c exists, which means that the limit of F(x) as x approaches c from both the left and right side are the same. 3. The limit is the same as the output. 
References
https://www.mathwarehouse.com/calculus/continuity/whataretypesofdiscontinuities.php
https://calculushowto.com/discontinuousfunction/
https://calcworkshop.com/limits/limitsandcontinuity/
http://scidiv.bellevuecollege.edu/dh/Calculus_all/CC_1_3_ContinuousFcns
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