# Differentiability Implies Continuity

An important theorem concerning derivatives is this:

If a function f is differentiable at x = a, then f is continuous at x = a.

The proof begins with the identity that for all $x\ne a$

$\displaystyle f\left( x \right)-f\left( a \right)=\left( {x-a} \right)\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}$

$\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\left( {f\left( x \right)-f\left( a \right)} \right)=\underset{{x\to a}}{\mathop{{\lim }}}\,\left( {\left( {x-a} \right)\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}} \right)=\underset{{x\to a}}{\mathop{{\lim }}}\,\left( {x-a} \right)\cdot \underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}$

$\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\left( {f\left( x \right)-f\left( a \right)} \right)=0\cdot {f}'\left( a \right)=0$

And therefore, $\underset{{x\to a}}{\mathop{{\lim }}}\,f\left( x \right)=f\left( a \right)$

Since both sides are finite, the function is continuous at x = a.

The converse of this theorem is false: A continuous function is not necessarily differentiable. A counterexample is the absolute value function which is continuous at the origin but not differentiable there. (The slope approaching from the left is not equal to the slope from the right.)

This is a theorem whose contrapositive is used as much as the theorem itself. The contrapositive is,

If a function is not continuous at a point, then it is not differentiable there.

Example 1: A function such as  $\displaystyle g\left( x \right)=\frac{{{{x}^{2}}-9}}{{x-3}}$ has a (removable) discontinuity at x = 3, but no value there.

So, in the limit definition of the derivative, $\displaystyle \text{ }\!\!~\!\!\text{ }\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{g\left( {3+h} \right)-g\left( 3 \right)}}{h}$ there is no value of g(3) to use, and the derivative does not exist.

Example 2:  $\displaystyle f\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {{{x}^{2}}} & {x\le 1} \\ {{{x}^{2}}+3} & {x>1} \end{array}} \right.$. This function has a jump discontinuity at x = 1.

Since the point (1, 1) is on the left part of the graph, if h > 0, $f\left( {1+h} \right)-f\left( 1 \right)>3$ and the limit  will always be a number greater than 3 divided by zero and will not exist. Therefore, even though the slopes from both side of x =1 approach the same value, namely 2, the derivative does not exist at x = 1.

This also applies to a situation like example 1 if f(3) were some value that did not fill in the hole in the graph.

On the AP Calculus exams students are often asked about the derivative of a function like those in the examples, and the lack of continuity should be an immediate clue that the derivative does not exist. See 2008 AB 6 (multiple-choice).

Just as important are questions in which the function is given as differentiable, but the student needs to know about continuity. Just remember: differentiability implies continuity. See 2013 AB 14 in which you must realize the since the function is given as differentiable at x = 1, it must be continuous there to solve the problem.

Continuity of the Derivative

A question that comes up is, if a function is differentiable is its derivative differentiable? The answer is no. While almost always the derivative is also differentiable, there is this counterexample:

$\displaystyle f\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {{{x}^{2}}\sin \left( {\frac{1}{x}} \right)} & {x\ne 0} \\ 0 & {x=0} \end{array}} \right.$

The first line of the function has a removable oscillating discontinuity at x = 0, but since the $\displaystyle {{x}^{2}}$ factor squeezes the function to the origin; the added condition that $\displaystyle f\left( 0 \right)=0$ makes the function continuous. Differentiating gives

$\displaystyle {{f}^{'}}\left( x \right)={{x}^{2}}\cos \left( {\frac{1}{x}} \right)\left( {\frac{{-1}}{{{{x}^{2}}}}} \right)+2x\sin \left( {\frac{1}{x}} \right)=-\cos \left( {\frac{1}{x}} \right)+2x\sin \left( {\frac{1}{x}} \right)$

And now there is no way to get around the oscillating discontinuity at x = 0.

# 2019 CED Unit 1 – Limits and Continuity

This is the first of a series of blog posts that I plan to write over the next few months, staying a little ahead of where you are so you can use anything you find useful in your planning. Look for this series every 2 – 4 weeks.

Unit 1 contains topics on Limits and Continuity. (CED – 2019 p. 36 – 50). These topics account for about 10 – 12% of questions on the AB exam and 4 – 7% of the BC questions.

Logically, limits come before continuity since limit is used to define continuity. Practically and historically, continuity comes first. Newton and Leibnitz did not have the concept of limit the way we use it today. It was in the early 1800’s that the epsilon-delta definition of limit was first given by Bolzano (whose work was overlooked) and then by Cauchy and Weierstrass. But their formulation did not use the word “limit”, rather the use was part of their definition of continuity. Only later was it pulled out as a separate concept and then returned to the definition of continuity as a previously defined term.

Students should have plenty of experience in their math courses before calculus with functions that are and are not continuous. They should know the names of the types of discontinuities – jump, removable, infinite, etc. As you go through this unit, you may want to quickly review these terms and concepts as they come up.

(As a general technique, rather than starting the year with a week or three of review – which the students need but will immediately forget again – be ready to review topics as they come up during the year as they are needed – you will have to do that anyway. See Getting Started #2)

### Topics 1.1 – 1.9: Limits

Topic 1.1: Suggests an introduction to calculus to give students a hint of what’s coming. See Getting Started #3

Topic 21.: Proper notation and multiple-representations of limits.

There is an exclusion statement noting that the delta-epsilon definition of limit is not tested on the exams, but you may include it if you wish. The epsilon-delta definition is not tested probably because it is too difficult to write good questions. Specifically, (1) the relationship for a linear function is always  $\delta =\frac{\varepsilon }{{\left| m \right|}}$  where m is the slope and is too complicated to compute for other functions, and (2) for a multiple-choice question the smallest answer must be correct. (Why?)

Topic 1.3: One-sided limits.

Topic 1.4: Estimating limits numerically and from tables.

Topic 1.5: Algebraic properties of limits.

Topic 1.6: Simplifying expressions to find their limits. This can and should be done along with learning the other concepts and procedures in this unit.

Topic 1.7: Selecting the proper procedure for finding a limit. The first step is always to substitute the value into the limit. If this comes out to be number than that is the limit. If not, then some manipulation may be required. This can and should be done along with learning the other concepts and procedures in this unit.

Topic 1.8: The Squeeze Theorem is mainly used to determine $\underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( x \right)}}{x}=1$ which in turn is used in finding the derivative of the sin(x). (See Why Radians?) Most of the other examples seem made up just for exercises and tests. (See 2019 AB 6(d)). Thus, important, but not too important.

Topic 1.9: Connecting multiple-representations of limit. This can and should be done along with learning the other concepts and procedures in this unit. Dominance, Topic 15, may be included here as well (EK LIM-2.D.5)

### Topics 1.10 – 1.16 Continuity

Topic 1.10: Here you can review the different types of discontinuities with examples and graphs.

Topic 1.11: The definition of continuity. The EK statement does not seem to use the three-hypotheses definition. However, for the limit to exist and for f(c) to exist, they must be real numbers (i.e. not infinite). This is tested often on the exams, so students should have practice with verifying that (all three parts of) the hypothesis are met and including this in their answers.

Topic 1.12: Continuity on an interval and which Elementary Functions are continuous for all real numbers.

Topic 1.13: Removable discontinuities and handing piecewise – defined functions

Topic 1.14: Vertical asymptotes and unbounded functions. Here be sure to explain the difference between limits “equal to infinity” and limits that do not exist (DNE). See Good Question 5: 1998 AB2/BC2.

Topic 1.15: Limits at infinity, or end behavior of a function. Horizontal asymptotes are the graphical manifestation of limits at infinity or negative infinity. Dominance is included here as well (EK LIM-2.D.5)

Topic 1.16: The Intermediate Value Theorem (IVT) is a major and important result of a function being continuous. This is perhaps the first Existence Theorem students encounter, so be sure to stop and explain what an existence theorem is.

The suggested number of 40 – 50 minute class periods is 22 – 23 for AB and 13 – 14 for BC. This includes time for testing etc. If time seems to be a problem you can probably combine topics 3 – 5, topics 6 -7, topics 11 – 12. Topics 6, 7, and 9 are used with all the limit work.

There are three other important limits that will be coming in later Units:

The definition of the derivative in Unit 2, topics 1 and 2

L’Hospital’s Rule in Unit 4, topic 7

The definition of the definite integral in Unit 6, topic 3.

Posts on Continuity

CONTINUITY To help understand limits it is a good idea to look at functions that are not continuous. Historically and practically, continuity should come before limits. On the other hand, the definition of continuity requires knowing about limits. So, I list continuity first. The modern definition of limit was part of Weierstrass’ definition of continuity.

Continuity (8-13-2012)

Continuity (8-21-2013) The definition of continuity.

Continuous Fun (10-13-2015) A fuller discussion of continuity and its definition

Right Answer – Wrong Question (9-4-2013) Is a function continuous even if it has a vertical asymptote?

Asymptotes (8-15-2012) The graphical manifestation of certain limits

Fun with Continuity (8-17-2012) the Diriclet function

Far Out! (10-31-2012) When the graph and dominance “disagree” From the Good Question series

Posts on Limits

Why Limits? (8-1-2012)

Deltas and Epsilons (8-3-2012) Why this topic is not tested on the AP Calculus Exams.

Finding Limits (8-4-2012) How to…

Limit of Composite Functions

Dominance (8-8-2012) See limits at infinity

Determining the Indeterminate (12-6-2015) Investigating an indeterminate form from a differential equation. From the Good Question series.

Locally Linear L’Hôpital (5-31-2013) Demonstrating L’Hôpital’s Rule (a/k/a L’Hospital’s Rule)

L’Hôpital’s Rules the Graph (6-5-2013)

Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.

2019 CED – Unit 1: Limits and Continuity

2019 CED – Unit 2: Differentiation: Definition and Fundamental Properties.

2019 CED – Unit 3: Differentiation: Composite , Implicit, and Inverse Functions

2019 CED – Unit 4 Contextual Applications of the Derivative  Consider teaching Unit 5 before Unit 4

2019 – CED Unit 5 Analytical Applications of Differentiation  Consider teaching Unit 5 before Unit 4

2019 – CED Unit 6 Integration and Accumulation of Change

2019 – CED Unit 7 Differential Equations  Consider teaching after Unit 8

2019 – CED Unit 8 Applications of Integration   Consider teaching after Unit 6, before Unit 7

2019 – CED Unit 9 Parametric Equations, Polar Coordinates, and Vector-Values Functions

2019 CED Unit 10 Infinite Sequences and Series

# How to Tell your Asymptote from a Hole in the Graph.

The fifth in the Graphing Calculator / Technology series

(The MPAC discussion will continue next week)

Seeing discontinuities on a graphing calculator is possible; but you need to know how a calculator graphs to do it. Here’s the story:

The number you choose for XMIN becomes the x-coordinate of the (center of) the pixels in the left most column of pixels. The number you choose for XMAX is the x-coordinate of the right most column of pixels. The distance between XMIN and XMAX is divided evenly between the remaining pixels so that all the pixels are evenly spaced across the screen (the same distance apart). The rows of pixels are done the same way evenly spacing them between YMIN and YMAX.

This spacing is usually not at “nice” values as can be seen by just moving the cursor across the screen and noticing the x-values or y-values at the bottom of the screen.

The cursor is located one pixel to the right of the y-axis and one pixel above the x-axis in the “standard” window of a TI-8x. Note the coordinates of that pixel at the bottom of the screen. These are the distances between the pixels.

To draw a graph, the calculator takes the x-coordinate of each pixel, calculates the corresponding y-value and turns on the pixel in that column with closest y-pixel-coordinate. If set in a connect mode, the calculator turns on several pixels in adjacent columns so that the y-values seem to connect; this is why the graph often looks jagged in steep sections of the graph. If you are in DOT mode, this does not happen and only one pixel in each column is on.

If you move the cursor over one of the points on a graph, you will see the pixel coordinates, NOT the actual y-coordinates. Use TRACE to see the actual y-coordinate. This is why when finding intersections, you should not just move the cursor over the point, but rather use “intersect” to see the actual y-value of the function.

If the function is undefined for some x-pixel value, then no pixel will turn on in that column. If the function is undefined for some value between the pixel values, then nothing happens because the calculator has not evaluated the function there, so the graph seems to be continuous.

Vertical “asymptotes” are the result of the calculator not evaluating the function at the undefined value; rather it connects the value on one side of the asymptote off the bottom of the screen with the next value on the other side of the asymptote off the top of the screen. If the asymptote appears exactly at a pixel value, then no “asymptote” will appear and that column of pixels will have no pixel turned on. (Some newer calculators and newer operating systems on older calculators have made adjustments so that the “asymptotes” do not show up. In some systems this feature can be turned on or off.)

The function $\displaystyle y=\frac{3\left( x-2 \right)}{\left( x-2 \right)\left( x+2 \right)}$ in the standard window. The vertical line is not really the asymptote and the “hole” at (2, 0.75) is not seen.

A removable discontinuity, a hole in the graph (really a skipped pixel), can be seen, if it occurs at a pixel value. Since in most examples the hole is at an integer or other “nice” number, you will not see them in the “standard” window. Use a “decimal” window, which has been chosen in advance so the x-values of the pixels are integers and nice decimals. (To see this, in a decimal window move the cursor around and notice the pixel coordinates).

The other thing you can do is adjust the XMIN and XMAX values so that the distance between them will land on integer values. (Nice project for your class – the number of pixels can be found in the guidebook, or you can count them. In the old days, before decimal windows, this was necessary – it was called finding a “friendly window.”)

The function $\displaystyle y=\frac{3\left( x-2 \right)}{\left( x-2 \right)\left( x+2 \right)}$ in the “decimal” window. The “asymptote” has disappeared and the “hole” at (2, 0.75) is now visible.

Zooming in or out may change these values so the hole or asymptote disappears.

For a related idea see the post My Favorite Function

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# Continuity 2

The definition of continuity of a function used in most first-year calculus textbooks reads something like this:

A function f is continuous at x = a if, and only if,

(1) f(a) exists (the value is a finite number),

(2) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists (the limit is a finite number), and

(3) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ (the limit equals the value).

A function is continuous on an interval if, and only if, it is continuous at all values of the interval. For the endpoints of closed intervals, the limits are adjusted to one-sided limits with x approaching a from inside the interval.

As I’ve written before, limits logically come before continuity since limits are used in the definition of continuity. But as a practical and historical matter continuity comes first. Continuity, or rather lack of continuity, gives us the examples that motivate the need for the concept of limit.

Karl Weierstrass (1815 – 1897) gave the modern definition of continuity: Given a function f and an element a of the domain I,   f is said to be continuous at the point a if for any number $\varepsilon >0$, however small, there exists a number $\delta >0$ such that for all x in the domain of f$\left| x-a \right|<\delta$ implies $\left| f\left( x \right)-f\left( a \right) \right|<\varepsilon$.

This looks very much like the definition of limit. In the delta-epsilon definition of limit the last inequality above is $\left| f\left( x \right)-L \right|<\varepsilon$ where L is the limit. Replacing the value with the limit allows a somewhat simpler wording of the definition of continuity than that given at the beginning, but adds the delta-epsilon complication. Weierstrass’ definition eliminates the need for saying the value and the limit are finite since that is assumed by writing f (a).

Some textbooks use the phrase “a function is continuous on its domain.” This seems somewhat limiting (no pun intended) in that a function such as $\displaystyle f\left( x \right)=\tfrac{1}{x}$ is certainly continuous on its domain but not continuous on the entire number line. We are usually concerned about where a function is not continuous, so first we find where it is not continuous: at the points not in its domain plus possibly other points in its domain.

Recent AP calculus exams (2012 AB4c, 2011 AB 6a) gave students a piecewise defined function and asked if it is continuous at the point where the two pieces meet. The question directed students to “use the definition of continuity to explain your answer” and “show that f is continuous.”  To answer the question students were expected to state what the two one-sided limits are and what the value there is. Since all these numbers are finite and equal the requirements of the definition are met.

This kind of question could be considered a question about continuity or a question about applying a definition (or theorem) to a particular situation. Either way students should understand the hypotheses of a definition or theorem and know how to verify that they are met in a particular situation.

# Continuity

Limits logically come before continuity since the definition of continuity requires using limits. But practically and historically, continuity comes first. The concept of a limit is used to explain the various kinds of discontinuities and asymptotes. Start by studying discontinuities.

Types of discontinuities to consider: removable (a gap or hole in the graph), jump, infinite (vertical asymptotes), oscillating, and end behavior (horizontal asymptotes).

Numerically: Make a table for the value of $\displaystyle \frac{1}{x-3}$ near x = 3 and as $\displaystyle x\to \pm \infty$. Relate the values and their signs to the graph. (Divide by a small number get a big number; divide by a big number, get a small number.)

Use the vocabulary of limits to explain the features of graphs. Example: The function $\displaystyle \frac{{{x}^{2}}-4}{x-2}$ has no value at x = 2 (f(2) does not exist), but as you get closer to x = 2 the function value gets closer to 4 ($\displaystyle \underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=4$).

Relate the limit, value and graph of the function. In the example above, the graph looks like the line $y=x+2$ with a gap or hole at the point (2, 4). Another example: $\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\frac{3{{x}^{2}}+x+8}{{{x}^{2}}+1}=3$ since, the graph gets closer to y = 3 as you go farther to the right. The line y = 3 is a horizontal asymptote.

Do this numerically as well:  $\displaystyle \frac{3{{x}^{2}}+x+8}{{{x}^{2}}+1}=3+\frac{x+5}{{{x}^{2}}+1}$ and since the fraction gets smaller as |x| gets larger, the function approaches 3 from above when x > 0 and from below when x < 0 (why?)

Extra for your experts: Discuss the reason for the jump discontinuity of

$\displaystyle f\left( x \right)=\frac{\cos \left( x \right)\sqrt{{{x}^{2}}-2x+1}}{x-1}$