1.4 Continuity

Continuity of a function at an endpoint

⊗ A left endpoint is continuous if lim[x→c⁺]f(x)=f(c) ⊗ A right endpoint is continuous if lim[x→c⁻]f(x)=f(c) To illustrate, look at the graph of the function f(x)=√x, which has the domain [0, ∞). Let's first examine a point in the interior of the domain, such as x = 1. Since f(1)=√1=1 and lim[x→1]√x=√1=1 the function f(x) is continuous at x=1. Now let's examine the endpoint to the left at (0,0). A two-sided would not make sense here since there is only one side to consider. Therefore, use a one-sided limit to describe continuity: f(0)=√0=0 and lim[x→0⁺]√x=√0=0 Since the one-sided limit and function are equal, we can say the function is continuous at its left endpoint. If you let c>0, you could test the function and limit values for all points in the domain (except the endpoint, but that has already been shown to be continuous): f(c)=√c and lim[x→c]√x=√c Since the function and limit values would be equal regardless of the value of c, you can say that the function is continuous at all points in its domain.

Summary of how to remove a removable discontinuity

1. Identify where a hole will exist in a graph 2. Evaluate the limit of a function at that point 3. Redefine the function so that the y-value at the hole is equal to the limit value (use a piecewise function)

Which three conditions could cause a function to be discontinuous at a point?

1. The function is undefined 2. The limit does not exist 3. The function and limit values are not equal Depending on which condition causes the discontinuity determines the type of discontinuity. If the limit fails to exist, why it does not exist is also important in determined in type of discontinuity.

Common types of discontinuities

1. removable (there is a hole in the graph) 2. jump (there is a jump in the graph) 3. infinity (there is a vertical asymptote in the graph) 4. oscillating (there is an infinite oscillation at a point in the graph)

Review Worksheet: Give a simple explanation of what it means for a function to be continuous.

A function is continuous if you can draw the whole graph without picking up your pencil. A more technical definition is that the right hand limit equals the left hand which equals the functions value at all points on its domain.

How can a limit fail to exist at a point?

A limit fails to exist at a point for one of three reasons: 1. the function is unbounded 2. the two one-sided limits are not equal to each other 3. the function oscillates between two values

Infinite Discontinuity

At least one of the one-sided limits is infinite, so the limit does not exist.The function value may or may not be defined (compare x = 1 and x = 2).

Review of Continuity Rollovers: How do you remove and removable discontinuity?

By defining the value of the function to be y-coordinate of the point that fills the hole.

How do you locate the position of a hole in the graph?

Determine the value for x where the hole exists; find the limit of the function as x approaches that value. This is the y-coordinate for the hole.

Continuity on an Open Interval Presentation

Identify the removable discontinuity in the function (sine(x+2))/(x+2). Redefine f(x) so that the function is continuous at that point. Since the function is a composition involving a trigonometric and rational functions, it is continuous on all points in its domain. This means the first step is to {identify where the function is undefined}. The function will only be undefined when the denominator is zero, which occurs when x=-2. Since f is not defined at x=-2, there is a discontinuity there. To determine if the discontinuity at x equals negative two is removable; evaluate the limit of the function as x→-2. If the limit exists, there is a removable discontinuity. We actually evaluated this limit expression using a graphical approach earlier in the unit. The limit of the function (sine(x+2))/(x+2) as x→-2 equals 1. Since the function is undefined at x=-2 but the limit exists there, we know there is a removable discontinuity at x=-2. Graphically, this means there is a hole in the graph of f at x=-2. The graph of the function f(x) confirms the location of the removable discontinuity. The limit value identifies the y-value of the hole in the graph of f, so we can use that value to redefine the function so that the discontinuity is removed. The easiest way to do this is with a piecewise function. In the first piece of the new function, use the original (sine(x+2))/(x+2) for all numbers except x equals negative two. For the second piece, use the limit's value for when x=-2.

How can you distinguish when a rational function has a hole or a vertical asymptote?

If there is a common factor in the numerator and denominator of a rational function, and x = c makes those factors equal to zero, there will be a hole at x = c. Any value for x that causes other factors in the denominator to be equal to zero cause vertical asymptotes.

Identifying Types of Discontinuity Bucket Game: lim[x→-2]f(x)=∞, lim[x→-2]f(x)=1, f(-2)=1

Infinite

Identifying Types of Discontinuity Bucket Game: lim[x→5]f(x)=0, lim[x→5]f(x)=-∞, f(5) is undefined

Infinite

Review Worksheet: {Part 3/4} There are four types on discontinuity. Sketch a graph of each and name them. Give an example of each.

Infinite Discontinuity When a function has infinite discontinuity, it has a vertical asymptote. The function approaches positive or negative infinity on one or both sides. This is the function f(x)={(-1/2)x+(1/2), x≤1 {1/(x-1), x>1

Identifying Types of Discontinuity Bucket Game: lim[x→0]f(x)=6, lim[x→0]f(x)=-2 f(0) is undefined

Jump

Identifying Types of Discontinuity Bucket Game: lim[x→2]f(x)=2, lim[x→2]f(x)=3, f(2)=2

Jump

Review Worksheet: {Part 2/4} There are four types on discontinuity. Sketch a graph of each and name them. Give an example of each.

Jump Discontinuity This is a function that has jump from one part of the function to the other. The right and left sided limits are not equal. This is the function f(x)={2x+1, x≤1 {-1, x>1

Review Worksheet: {Part 4/4} There are four types on discontinuity. Sketch a graph of each and name them. Give an example of each.

Oscillating Discontinuity Oscillating discontinuity is when a function oscillates back and forth infinitely without approaching a specific value. This is the function f(x)=cos(1/(x-1))+1

Identifying Types of Discontinuity Bucket Game: lim[x→1]f(x)=4, lim[x→1]f(x)=4, f(1)=7

Removable

Identifying Types of Discontinuity Bucket Game: lim[x→3]f(x)=0, lim[x→3]f(x)=0, f(3) is undefined

Removable

Review Worksheet: {Part 1/4} There are four types on discontinuity. Sketch a graph of each and name them. Give an example of each.

Removable Discontinuity A function has a removable discontinuity when there is a hole. To remove the discontinuity, redefine the function as a piecewise function and 'fill in' the hole (See #3 and 4) This is the function f(x)=(x³-x-2)/(x-2).

Example: Removable Discontinuity Let's look at an example of how to redefine a function so as to eliminate a removable discontinuity. Identify the discontinuities in the graph of f(x)=(3x-6)/(x²-6x+8). If there is a removable discontinuity in the graph, redefine the function to make it continuous at that point.

Start by factoring the rational expression, which results in the common factor of (x-2) in the numerator and denominator: f(x)=(3x-6)/(x²-6x+8) f(x)=(3(x-2))/((x-4)(x-2)) Since x=4 would cause a vertical asymptote in the graph of f(x), there must be an infinite discontinuity at x=4. Because of the common factor of (x-2) in the numerator and denominator, there will be a hole in the graph when x=2. This is a removable discontinuity. In order to remove the discontinuity at x=2 (in other words, to fill in the hole), find the limit at x=2 lim[x→2](3x-6)/(x²-6x+8) =lim[x→2]3/(x-4) =3/((2)-4) =-3/2 We now know the hole is at the coordinate (2, -3/2). If we redefine the function f(x) to be equal to -3/2 when x=2, the discontinuity will be removed. To redefine the function, use a piecewise function where the top piece is the original equation except at x=2 and the bottom piece is the value that fills the hole: f(x)= {(3x-6)/(x²-6x+8), x≠2 {-3/2 x=2 Another way to redefine a rational function to eliminate a hole is to reduce the common factors. In this example, the redefined function f(x)=3/(x-4) fills in the hole in the graph at x=2 just as the piecewise function did. It is important to realize that the original f(x) function (shown in figure 1) and the redefined f(x) function (figure 2) are not the same function. These two functions differ when x=2.

Review of Continuity Rollovers: Where is the discontinuity and why is it nonremovable?

The discontinuity is at x=2. The discontinuity is nonremovable because f(x) does not approach the same value from the right and the left at x=2. There is a "jump."

Review of Continuity Rollovers: If f has a nonremovable discontinuity at c because f(x) is unbounded, what pattern will you see in your graph?

The function will start to resemble a vertical line near c. There will be a "break" in the graph.

Review of Continuity Rollovers: Does the graph have a discontinuity? If so, where? Is it removable or non removable? Justify your answer.

The graph has a discontinuity at x=-2. Since f(x) is unbounded at x=-2 (there is a break in the graph), the discontinuity is nonremovable.

Review of Continuity Rollovers: Does the graph have a discontinuity? If so, where? Is it removable or nonremovable? Justify your answer.

The graph has a discontinuity at x=2. Since f(x) is oscillating at x=2 (there is a smudge on the graph), the discontinuity is nonremovable.

Review of Continuity Rollovers: How can you tell by looking at a graph that discontinuity is removable?

The graph will have only a hole (not a jump, break, or smudge) because in order for a discontinuity to be removable, the function must approach the same value from the right and the left.

Oscillating

The limit does not exist because of an infinite oscillation at a single point.The function value may or may not be defined (compare x = 1 and x = 2).

Removable Discontinuity

The limit exists, but either the function is undefined (see x = 1) or does not equal the limit value (see x= 2). Occurs when there is a hole in the graph at a point. The function value may or may not be defined at that point. If the function value is defined, however, it cannot be the same value as the limit value (otherwise, it could be continuous!).

Jump Discontinuity

The two one-sided limits are not equal, so the limit does not exist. The function value may or may not be defined (compare x = 1 and x = 2).

Review Worksheet: Can you make the function f(x)=(2x+6)/(x²+5x+6)

This function is not continuous at x=-3 or x=-2. See if the function has limit at either of these values. lim[x→-3]=(2x+6)/(x²+5x+6) =(2(x+3))/((x+3)(x+2)) =2/(x+2) lim[x→-3]2/(x+2) =2/((-3)+2) =2/-1 =-2 lim[x→-3]=-2 lim[x→-2]=(2x+6)/(x²+5x+6) =(2(x+3))/((x+3)(x+2)) =2/(x+2) lim[x→-2]=2/(x+2) =2/((-2)+2) =2/0 =DNE We can remove the discontinuity at x=-3 by making f(-3)=-2. The discontinuity at x=-2 is infinite, therefore it cannot be removed.

Review Worksheet: Can you make the function f(x)=(x²-4x-5)/(x-5) continuous?

This function is not continuous at x=5. See if the function has a limit at x=5. lim[x→5](x²-4x-5)/(x-5) =((x-5)(x+1))/(x-5) =(x+1)/1 =(x+1) lim[x→5](x+1) ((5)+1) =6 lim[x→5]=6 Redefine the function using the piecewise function f(x)={(x²-4x-5)/(x-5), x≠5 {6, x=5

Evaluate the limit as x→4 for the piecewise function: f(x)= {2x-1, x<4 {5+√x, x>4

Use the top piece for the left-sided limit and the bottom piece for the right-sided limit: lim[x→4⁻]f(x) =lim[x→4⁻](2x-1) =2(4)-1 =7 lim[x→4⁺]f(x) =lim[x→4⁺](5+√x) =5+√4 =7 Since both one-sided limits equal 7, f (x) → 7 as x → 4.

Point of discontinuity

When there are points in the domain where a function is not continuous (or discontinuous), the function is said to have a point of discontinuity.

Concept of continuity

an important quality of functions involving calculus. relies heavily on limits

Rewritten Steps≠

f(x)=(sine(x+2))/(x+2) Procedure: Find when the function is undefined. f(x) is undefined at x=-2 Evaluate the limit at the point where the function is undefined. lim[x→-2](sine(x+2))/(x+2)=1 Since the limit exists where the function is undefined, there is a removable discontinuity. Now redefine the function using the limit value to remove the discontinuity. Redefined Function: f(x)={(sine(x+2))/(x+2), x≠-2 {1, x=-2

Continuous function

when the function value and limit value are equal at the same point, the function is described as continuous at that point. More formally, a function f(x) is continuous at a point, c if: 1. f(c) is defined 2. the lim f(x) (as x approaches c) exists 3. the lim f(x) (as x approaches c) = f(c) A function that is continuous at all points in its domain is said to be a continuous function. Most common function-polynomial, rational, radical, exponential, logarithmic, and trigonometric-are continuous in their domain. Don't forget that some of these have holes and vertical asymptotes in their graphs, but those x-values are not in their domain.

Continuity Review Game: Match the discontinuity with the correct function) {4} oscillating discontinuity at x=1

{a} f(x)=cos(1/x)-1 The graph of f(x) shows an infinite oscillation at x=1.

Continuity Review Game: Match the discontinuity with the correct function) {5} continuity for all x

{b} f(x) is a piecewise function defined by 2^x when x<1 and by 1+x² when x≥1 f(x)→2 as x→1, but f(1)=2. Since the function pieces are exponential and polynomial, f(x) is continuous everywhere else.

Continuity Review Game: Match the discontinuity with the correct function) {2} jump discontinuity at x=1

{c} f(x) is a piecewise function defined by 5x² when x≤1 and by (√x+3) when x>1. f(x)→5 as x→1⁻, f(x)→2 as x→1⁺

Continuity Review Game: Match the discontinuity with the correct function) {7} jump discontinuity at x=-1

{d} f(x)=2|x+1|/(x+1) The graph jumps from y=-2 to y=2 at x=-1

Continuity Review Game: Match the discontinuity with the correct function) {8} infinite discontinuity at x=-1

{e} f(x)=(2x²-x-1)/(x+1) The graph has a vertical asymptote at x=-1

Continuity Review Game: Match the discontinuity with the correct function) {1} removable discontinuity at x=1

{f} f(x) is the piecewise function defined by 2x-5 when x<1 and x²-4 when x>1. f(x)→-3 as x→1, but f(1) is undefined

Continuity Review Game: Match the discontinuity with the correct function) {6} removable discontinuity at x=-1

{g} f(x)=(2x²+x-1)/(x+1) The (x+1) factor is common to the numerator and denominator, so there is a hole at x=-1.

Continuity Review Game: Match the discontinuity with the correct function) {3} infinite discontinuity at x=1

{h} f(x)=log(x-1) The graph of f(x) has a vertical asymptote at x=1.