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# 4.1: Functions, Limits, and Continuity

Difficulty Level: At Grade Created by: CK-12

## Equations and Graphs

CONTENT

Students that are strong in algebra and motivated will appreciate a discussion of graphs/equations that consists essentially of examples like \begin{align*}y = x, y = x^2, y = x^n, x^2 + y^2 = R, xy = c, x^2 - y^2 = c, \frac{x^2}{a^z} + \frac{y^z}{b^z} = 1, y =\mathrm {Sin} (x),\end{align*} and so on. The plots could be shown and different characteristics of each could be discussed.

Students who are not as strong in algebra or are not as motivated may be better served by first reviewing rectangular coordinates, and then drawing some plots like a circle or a parabola. It can be pointed out that each point in any picture always has an \begin{align*}x\end{align*} and a \begin{align*}y\end{align*} coordinate. So we could prescribe how to draw a picture by listing the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} values that should be darkened. An equation is just this, a description of which \begin{align*}x\end{align*} and \begin{align*}y\end{align*} values should contain ink and which should not.

PROCESS

The less experienced or motivated students should see the content of this lesson fine-tuned to their level. Equations and graphs may be introduced gradually by drawing a rectangular coordinate grid and going through Cartesian Coordinates as a means of locating points. They could be motivated by saying something like, suppose a person needs to a buried treasure. If the treasure is at the coordinates \begin{align*}(3,-7)\end{align*} then where is it? And so on.

Then, a simple equation like \begin{align*}y=x+2\end{align*} can be introduced by asking the question, “Which points \begin{align*}(x,y)\end{align*} on the graph I’ve drawn will make this equation true?” By taking some incorrect and some correct guesses, we should eventually arrive at the line:

One can then progress to more and more complicated equations like \begin{align*}y=x^2\end{align*} and \begin{align*}y=\sqrt{x-1}\end{align*}.

The more advanced or motivated student should be immersed in problem solving related to this lesson. For example, an advanced student might be asked to formulate an equation that describes some process, and then to plot this equation as a graph. Here are some possibilities:

Question 1:

You are given \begin{align*}\20\end{align*} each Monday morning as an allowance. This will allow you to spend \begin{align*}x\end{align*} dollars each weekday for lunch and \begin{align*}y\end{align*} dollars over the weekend so that you use it all up by Monday. Find an equation involving \begin{align*}x\end{align*} and \begin{align*}y\end{align*} to describe this and then plot the possible values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*} that solve this equation.

\begin{align*}20-5x-y=0\end{align*} or equivalently \begin{align*}y=20-5x\end{align*}. And the plot should look like:

Question 2:

A right triangle with height \begin{align*}x\end{align*} and width \begin{align*}y\end{align*} has area equal to \begin{align*}100\end{align*}. Find an equation to describe the possible values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*} and plot the values that solve it.

Answer 2: \begin{align*}\frac{xy}{2} = 100\end{align*} or equivalently \begin{align*}y = \frac{200}{x}\end{align*}. And the plot should look like:

Question 3:

If y is the area of a square and \begin{align*}x\end{align*} is the perimeter, what is an equation relating \begin{align*}y\end{align*} and \begin{align*}x\end{align*} and a plot of the values that solves the equation.

Answer 3: \begin{align*}y = \left (\frac{x}{4} \right )^2\end{align*} or equivalently \begin{align*}y = \frac{1}{16} x^2\end{align*}. And the plot should look like:

PRODUCTS

There are a variety of different ways in which a student can demonstrate mastery of equations and graphs. The most straightforward are simply drawing graphs for a variety of different equations or looking at graphs and recognizing the corresponding equations. This could be done, for example with a matching game where graphs are on the right and equations are on the left and the object is to decide which goes with which.

Another possibility is to assign a complicated equation to each student like \begin{align*}y = e^{\mathrm {Sin}(x^2)}\end{align*} or may be \begin{align*}y = \mathrm {Cos} \left (\frac{1}{x} \right )\end{align*} and ask them to graph their equation carefully with a description of the properties, like what happens as \begin{align*}x\end{align*} becomes very large or very small, or negative. Then each student could give a brief presentation of his/her equation on the board in which s/he would draw the graph and discuss why the equation implies certain properties of the graph.

## Relations and Functions

CONTENT

The concept of a function has a very natural splitting into two levels. The content for students who are very motivated and enjoy math could begin with the most general description of a function in terms of sets. A function \begin{align*}f\end{align*} in its most abstract sense is just a set of ordered pairs of elements from two sets \begin{align*}A\end{align*} and \begin{align*}B\end{align*}. That is, \begin{align*}f=\left \{(a,b) | a \in A,b \in B \right \}\end{align*}. The set \begin{align*}A\end{align*} is called the domain of \begin{align*}f\end{align*} and the set \begin{align*}B\end{align*} is called the range. In order to think of a function as a machine taking elements of \begin{align*}A\end{align*} and producing elements of \begin{align*}B\end{align*}, we should require that each \begin{align*} a \in A\end{align*} belongs to only pair in \begin{align*}f\end{align*}. That way, each \begin{align*}a\end{align*} is sent to only one element of \begin{align*}B\end{align*}.

Then this concept could be specified to the particular nature of single-variable calculus. In this subject functions map numbers to numbers, and the domain and range are just open or closed subsets of the real number line \begin{align*}\mathbb{R}=(- \infty , \infty)\end{align*}.

For the less motivated or experienced student, this content should be made more geometric by focusing on the vertical-line test for functions. A graph is the graph of a function if any vertical line intersects the curve only once (or not at all). The domain of a function is the shadow cast upon the \begin{align*}x-\end{align*}axis by a light from above and the range is the shadow cast on the \begin{align*}y-\end{align*}axis by a light from the right.

PROCESS

The learning process here could take many different routes. It is a good idea to begin with students’ intuitions, and so for a first activity it would be good to discuss what an idealized machine does. For example, consider the following function:

Suppose we pick a thesaurus up, and use it as a machine for turning words into other words. Given some word like, say, “happy”, we look up the word and choose the first synonym that appears. If we were using thesaurus.com then we would have in this case that: \begin{align*}f\end{align*} (happy)=blessed.

Then there are a variety of function characteristics that can be seen in this example. We note that not all words appear in a thesaurus, so the domain is limited and so is the range. In fact, the domain and range are probably just about the same subsets of the English language.

Furthermore, we see that using just one thesaurus we always get just one result from a given word. It is not possible for \begin{align*}f\end{align*}(happy)=blessed and simultaneously to have that \begin{align*}f\end{align*}(happy)=content. This means that for a given input, we will have one fixed output. On the other hand, we might get the result “blessed” for a different input as well. This means that if \begin{align*}f(a)=c\end{align*} the we may also have that \begin{align*}f(b)=c\end{align*} as well.

Another fun project is for students to think of as many sets (or categories) as possible and write these on the board, like:

{brands of cereal},{famous people},{colors}, {songs},{movies},{whole numbers} and so on. Then the class could split into partners and each duo could be tasked with determining a function from one of these sets to another. The pair should carefully describe the domain and range, and why it is that this is indeed a function. In other words, why is it that no object from one set is mapped to more than one from another. As an example from the sets above, consider {brands of cereal} and {famous people}. We could have a function that takes a famous person and maps him/her to his/her favorite brand of cereal. In order to be sure this is a function we must include only famous people who have a single favorite brand. If they have more than one, that would violate our rule of functions. Furthermore, famous people with no favorite brand of cereal do not belong to the domain and cereals that nobody likes would not be in the range.

This would be a good subject to learn using small groups as well. The class could be split into groups of three and each group could be given an equation like \begin{align*}y^2+x^2=4\end{align*} or like \begin{align*}y = \mathrm {Sin}(x)\end{align*} and the group would first graph the equation. Then they would have to determine if it is a function and then figure out the domain and range if it is a function. This could be presented to the class afterwards. Each group should be closely monitored to make sure that each member is participating and in order to be sure of this fact, it would be advisable to have members each produce paperwork describing the qualities of the group’s equation.

PRODUCTS

There are many possible ways to test a student’s knowledge of this material. The most basic is to provide a series of graphs and ask which are graphs of functions, and a series of equations and ask which are equations representing functions with input \begin{align*}x\end{align*} and output \begin{align*}y\end{align*}. The follow-up, of course, asks in the case that the equation or graph is not a function then what \begin{align*}x\end{align*} value is mapped to more than one \begin{align*}y\end{align*} and in the case that it is a function, what are the domain and range.

The nice thing about these problems is that they are simple to come up with in large quantity, can be tailored to the level of any student, and will introduce important equations that will be used later. However, for the student that is very difficult to motivate it may be better to use examples from everyday life. For example, consider the “Wal-Mart Function” which takes a product and returns the price. Why is this a function? Or perhaps, what do we have to specify in order to make sure that it is a function? What is the domain and what is the range? A great variety of different functions like this can be considered, and thinking of more everyday things may motivate students that do not typically enjoy math. Here are some more examples:

1. Wal-Mart Function: \begin{align*}\mathrm{products} \rightarrow \mathrm{price}\end{align*} For example: \begin{align*}f(\mathrm{dollhouse})= \149.99\end{align*}
2. Mother’s Name Function: \begin{align*}\mathrm{people} \rightarrow \mathrm{mother' s name}\end{align*}
3. Inverse Mother’s Name Function: \begin{align*}\mathrm{mother's name} \rightarrow \mathrm{person}\end{align*} (Is this a function??)
4. Mountain Height Function: \begin{align*}\mathrm{mountain name} \rightarrow \mathrm{height above sea level}\end{align*}

## Models and Data

CONTENT

The content in this lesson begins with a simple data set containing a list of \begin{align*}x-\end{align*}values and the corresponding \begin{align*}y-\end{align*}values like:

\begin{align*}& x && y \\ & 0 && 0 \\ & 1 && 10 \\ & 2 && 30 \\ & 3 && 5 \\ & 4 && -7 \\ & 5 && -205 \\ & 6 && 3.141\end{align*}

This content cannot be differentiated in any way although it should be said that a more basic approach for the less-experienced should be more concrete. The numbers should be given explicitly and should represent something concrete like the profit for a given year. A more advanced or motivated student might appreciate abstraction where the data is given as simply a set of n number pairs: \begin{align*}\left \{(x_i,y_i)|i=1,2, \ldots ,n \right \}\end{align*}.

The functions to which one seeks to fit this data have progressive levels of difficulty. And this gives an opportunity for differentiation. A more basic student may be taught only to fit data only to functions of the form \begin{align*}y = a + bx\end{align*} or \begin{align*}y = a + bx + cx^2\end{align*}. More generally, the most basic fits correspond to polynomial fits of the form: \begin{align*}y = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots\end{align*}. It turns out to be the case that any data set containing \begin{align*}n\end{align*} points can be exactly fit to a polynomial with highest power \begin{align*}n-1\end{align*}.

The more motivated or experienced student can also try fits with trigonometric functions of the form \begin{align*}y = a \mathrm {Sin}(bx+c)\end{align*} or with exponential functions of the form \begin{align*}y = a \ b^{cx} + d\end{align*}. It can be pointed out that these are actually like fitting the data to an infinitely long polynomial. If this piques interest then students can be assured that they will learn more about this towards the end of their calculus class when infinite series are discussed.

PROCESS

The process of fitting data presented in this chapter leaves little room for differentiation since it is done entirely by calculator. A fun homework project for students may be to find some data sets on the internet and then do some fitting.

A fun project would be to predict the temperature for the next \begin{align*}2\end{align*} or \begin{align*}3\end{align*} days by using temperature data from the past and a reasonable fit. The students could look up past values, create a fit, and then project the fit beyond the present to see what will happen. This could even be done with some past date. The teacher could bring in temperature data for \begin{align*}100\end{align*} consecutive days in \begin{align*}1984\end{align*} and have students fit this to a function. There could be a competition to see which student is best able to predict the future.

This could also be done with stock prices, something that may be more exciting since it involves money. The teacher could present the price of a certain stock for the past \begin{align*}100\end{align*} trading days and students could be divided into groups. Each group could be asked to try a different kind of fit for this data, and then use this to predict what will happen over the next ten days. Alternatively, students could try various plots and decide which to believe. There could even be a virtual stock market where students are each student virtually purchases \begin{align*}10\end{align*} stocks and must decide based upon their fits whether to buy or sell more stock from classmates.

PRODUCTS

As mentioned in the above, it would be a fun project to have students use data to predict the future of a stock price or the weather. This it seems would be the most engaging end product for all of the students to produce since it is so tangible.

On the other hand, there is always the more abstract option which would be to provide lists of data and ask for the plot variables \begin{align*}a,b,c\end{align*},etc. that best fit this to a certain form. The student could be asked how good this plot is by using numbers given by most calculators like \begin{align*}R-\end{align*}values.

## The Calculus

CONTENT

This material can be presented to the entire class by focusing carefully on the geometric nature of every idea. Previously, the students have understood how to think about the steepness of a straight line through the points \begin{align*}(x_1,y_1 )\end{align*} and \begin{align*}(x_2,y_2)\end{align*}. This is just the slope \begin{align*}m=\frac{y_2-y_1}{x_2-x_1}\end{align*}.

However, if the graph is not straight but curved between these points then what does the number \begin{align*}m=\frac{y_2-y_1}{x_2-x_1}\end{align*} represent geometrically. It isn’t the slope or steepness at \begin{align*}(x_1,y_1)\end{align*} or at \begin{align*}(x_2,y_2)\end{align*}. It’s sort of like the average steepness between these two points. So how would we find the steepness at just one point? Well, by drawing a picture like this:

it should become clear that the slope of the red lines gets closer and closer to slope at p as we move the two points nearer and nearer.

The line whose slope we are examining with each guess always passes through two points and is called a secant. But with each guess the secant line becomes closer and closer to a line which passes through just one point. This is the tangent line, and if we zoom way in on the point \begin{align*}p\end{align*} then the curve itself and the tangent will be indistinguishable. That is why we often refer to the tangent as a linear approximation to the curve at \begin{align*}p\end{align*}.

Connecting this idea of tangents with areas in a casual or qualitative way is challenging. However, the geometric connection can described if we let \begin{align*}f(x)\end{align*} be any function and \begin{align*}F(z)\end{align*} be the function that gives the area under \begin{align*}f(x)\end{align*} between \begin{align*}x=0\end{align*} and \begin{align*}x=z\end{align*}:

Then \begin{align*}F(z)\end{align*} is not changing only when \begin{align*}f(z)=0\end{align*}, and is increasing at a rate that is proportional to \begin{align*}f(z)\end{align*}. That is, the bigger \begin{align*}f(z)\end{align*} is at any point, the faster the area is increasing as we move to the right (that is, as \begin{align*}z\end{align*} increases).

Now, the rate of change for a function like \begin{align*}F(z)\end{align*} at a point z is its steepness or slope at the point \begin{align*}z\end{align*}. So we see that the rate of change for the area function \begin{align*}F(z)\end{align*} is equal to the function whose area it is describing \begin{align*}f(z)\end{align*}. This is the content of the Fundamental Theorem of Calculus.

PROCESS

Getting students to understand these ideas can be tricky. There are a number of approaches that could be tried. The first would be to draw a big Cartesian coordinate system on the board with a big blown up version of \begin{align*}y=x^2\end{align*}. Then select two points like \begin{align*}(1,1)\end{align*} and \begin{align*}(4,16)\end{align*} and have a student come draw the line that connects these points. Then have the class calculate this line’s slope. Next do the same for the points \begin{align*}(1,1)\end{align*} and \begin{align*}(3,9)\end{align*} and then for \begin{align*}(1,1)\end{align*} and \begin{align*}(2,4)\end{align*} and then maybe for \begin{align*}(1,1)\end{align*} and \begin{align*}(1.414,2)\end{align*}. Point out how the lines that were drawn are getting closer and closer to the tangent line, and that the slopes are getting closer and closer to the number \begin{align*}2\end{align*}.

Another great technique involves a fun little riddle. The average speed for a trip is the total distance traveled divided by the total time. This is like the secant line for the position plot. Suppose that a racecar is planning two laps around a \begin{align*}2 \;\mathrm{mile}\end{align*} track. If averages \begin{align*}60\frac{\mathrm{mi}}{\mathrm{hr}}\end{align*} over the first lap, how fast must it travel over the second lap to average \begin{align*}180 \frac{\mathrm{mi}}{\mathrm{hr}}\end{align*} over both miles combined?

Now, speed is distance over time so we want to think of distance as the \begin{align*}y-\end{align*}variable and time as the \begin{align*}x-\end{align*}variable. At an average speed of \begin{align*}60\end{align*}, the first mile will take one minute. So we know the car’s plot over the first mile will look something like:

where we actually don’t know the shape of the curve between \begin{align*}(0,0)\end{align*} and \begin{align*}(1,1)\end{align*}. All we know is that it does make it to this point. In order to average \begin{align*}120 \;\mathrm{mi/hr}\end{align*} overall, the car would have to travel two miles in \begin{align*}1 \;\mathrm{minute}\end{align*}. So the curve would have to pass through the point indicated here as well:

However, there is clearly no way to continue the curve in the first plot through the point in the second plot without going vertically at some point. This would mean the steepness would be undefined, or infinite, and this is not a valid speed. Therefore it is impossible to average \begin{align*}180 \;\mathrm{mi/hr}\end{align*} overall after averaging \begin{align*}60 \;\mathrm{mi/hr}\end{align*} for a mile. The minute needed has already been used up.

PRODUCTS

This material is difficult to test since it is largely qualitative. However, questions asking for a written or verbal description of the derivative or tangent line would work. Students could be divided into small groups and asked to approximate some quantity like the slope of the tangent line to \begin{align*}y=3x^2-4x+5\end{align*} at \begin{align*}x=2\end{align*} or the area under \begin{align*}y=3x^2-4x+5\end{align*} between \begin{align*}x=0\end{align*} and \begin{align*}x=3\end{align*}. They should accomplish each of these tasks using successive approximations as secant lines or Riemann rectangles respectively.

## Finding Limits

CONTENT

Note: It is strongly encouraged that teachers use the variable \begin{align*}h\end{align*} as the independent variable in this lesson instead of \begin{align*}x\end{align*} as the author of the text has used. This will significantly reduce the burden on students when both \begin{align*}x\end{align*} and \begin{align*}h\end{align*} are involved and limits are being taken for \begin{align*}h\end{align*} only.

In order to make this material accessible to anyone, the introduction should be as intuitive as possible. For example, in the last lesson we described that finding the slope at a point or the area under a curve involves a process of taking better and better estimates. So the question is: What do we do if instead of having an exact number, all we have is a bunch of progressive estimates for it?

Recall that in the last lesson we were estimating the slope of \begin{align*}y=x^2\end{align*} at \begin{align*}x=1\end{align*}, and perhaps if we had continued this process we’d get a series of estimates that look like: \begin{align*}\left \{6.92,4.23,2.1,2.006,2.00007,\ldots \right \}\end{align*}. We may even be able to write these as a function like \begin{align*}f(h)= \frac{1}{h} ((1+h)^2-1)\end{align*} where these are just successive values as we let \begin{align*}h\end{align*} get closer and closer to zero. Now, while it could be clear that the estimates are getting closer and closer to the number \begin{align*}2\end{align*}, they also never actually get there. And we can’t very well just plug \begin{align*}h=0\end{align*} into the function since there is an \begin{align*}h\end{align*} in the denominator and we don’t know how to divide by \begin{align*}h\end{align*}.

For this reason we must define a concept of limit so that we can concretely say that

\begin{align*}\lim_{h \to 0} \frac{1}{h} ((1 +h)^2 - 1) = 2\end{align*}

This is no simple task since, as we said, \begin{align*}h\end{align*} cannot just be plugged in directly. Despite this, we can easily see by using our calculator or a piece of paper that choosing smaller and smaller values of \begin{align*}h\end{align*} gives results that are closer and closer to \begin{align*}2\end{align*}. So maybe we can have some way of saying mathematically that the values get closer and closer to \begin{align*}2\end{align*} as \begin{align*}h\end{align*} gets smaller.

This is the content of the formal definition for a limit’s existence. We first need to specify the phrase closer and closer. In order to make this concrete, we say that for any distance you can think of, no matter how small, we will eventually be even closer than that! Furthermore, we will stay that close or closer for better and better approximations.

The statement in the book for the result that:

\begin{align*}\lim_{h \to 0} \frac{1}{h} ((1 +h)^2 - 1) = 2\end{align*}

can be translated as follows: Pick the smallest positive distance you can think of. Then I can find a distance so that if \begin{align*}h\end{align*} is less than that distance from \begin{align*}0, \frac{1}{h} ((1+h)^2-1)\end{align*} will be less than your distance from \begin{align*}2\end{align*}. That is, no matter how close to \begin{align*}2\end{align*} you want me to get I can find \begin{align*}h-\end{align*}values that will get the function within that distance.

PROCESS

This is really the first very conceptually challenging concept that most students learn in all of math. And indeed, many students will simply leave the definition of a limit behind without fully getting it. It may even be this concept that plays a large role in whether a student will like calculus or hate it. So the teacher’s responsibility to convey this idea intuitively and clearly is a great one.

The only real way for students to get this definition is by trying to formulate it themselves. Begin with the simple function:

\begin{align*}f(h) = \frac{h^2 - 1}{h - 1}\end{align*}

Now, clearly \begin{align*}f(1)\end{align*} is not defined since there’s a zero in the denominator. On the other hand you can probably see that \begin{align*}f(h)\end{align*} can be simplified as \begin{align*}f(h)=h+1\end{align*} but this is only valid when \begin{align*}h\end{align*} is not equal to \begin{align*}1\end{align*}. Then pose the question: How could we say that as \begin{align*}h\end{align*} gets closer and closer to \begin{align*}1, f(h)\end{align*} gets closer and closer to \begin{align*}2\end{align*}?

The class should be divided into small groups where each group ideally contains at least one member who is strongly motivated mathematically. This will help motivate the other members. Then, the group should come up with a precise description of the behavior that \begin{align*}f(h)\end{align*} is getting closer to \begin{align*}2\end{align*} as \begin{align*}h\end{align*} is getting closer to \begin{align*}1\end{align*}. The group will have succeeded if the teacher:

1. Can show that according to the group’s precise description, \begin{align*}f(h)\end{align*} is getting closer to \begin{align*}2\end{align*} as \begin{align*}h\end{align*} gets closer to \begin{align*}1\end{align*}
2. \begin{align*}f(h)\end{align*} is NOT getting closer to something other than \begin{align*}2\end{align*} as \begin{align*}h\end{align*} gets closer to \begin{align*}1\end{align*}
3. \begin{align*}f(h)\end{align*} is NOT getting closer to \begin{align*}2\end{align*} as \begin{align*}h\end{align*} gets closer to something other than \begin{align*}1\end{align*}

Clearly, the teacher’s job here is quite difficult. S/he must move from group to group assisting with the thought process and gently nudging the group towards a definition like the one in the book. If the students are delicately allowed to struggle with these ideas, and to talk them over carefully, then the definition will become clear.

PRODUCTS

Clearly all of the students will have difficulty grasping the definition of a limit’s existence. It is therefore recommended that for students that are having significant difficulty, problems are given which only involve steps towards understanding limits. For example, they could numerically find limits of various complicated with canceling factors like:

\begin{align*}\lim_{h \to 1} \frac{5 - 4h - h^2 - 3h^3 + 3h^4 - h^5 + h^6}{-24 + 24h + h^2 + h^4 + 3h^7 - 3h^8}\end{align*}

or the famous example:

\begin{align*}\lim_{h \to 0} \frac{\text {Sin}(x)}{x}\end{align*}

1. How small must we choose \begin{align*}h\end{align*} so that \begin{align*}f(h) = \frac{h^2 -1}{h - 1}\end{align*} is between \begin{align*}1.9\end{align*} and \begin{align*}2.1\end{align*}
2. How small must we choose \begin{align*}h\end{align*} so that \begin{align*}f(h) = \frac{h^2 -1}{h - 1}\end{align*} is between \begin{align*}1.99\end{align*} and \begin{align*}2.01\end{align*}
3. How small must we choose \begin{align*}h\end{align*} so that \begin{align*}f(h) = \frac{h^2 -1}{h - 1}\end{align*} is between \begin{align*}1.999\end{align*} and \begin{align*}2.001\end{align*}
4. How small must we choose \begin{align*}h\end{align*} so that \begin{align*}f(h) = \frac{h^2 -1}{h - 1}\end{align*} is between \begin{align*}2 - \in\end{align*} and \begin{align*}2 + \in\end{align*}

The more advanced student could be challenged directly to find deleted neighborhoods of the limit point \begin{align*}a\end{align*} (for \begin{align*}x \rightarrow a\end{align*}) which correspond to given neighborhood \begin{align*}D\end{align*} of a limit \begin{align*}L\end{align*}. That is, for example:

1. Show directly using the definition of a limit that \begin{align*}\lim_{h \to 0} \frac{h^2 - 1}{h - 1} = 2\end{align*}
2. Show directly using the definition of a limit that \begin{align*}\lim_{h \to 1} \frac{h^2 - 4h + 3}{h^2 + h - 2} = - \frac{2}{3}\end{align*}

## Evaluating Limits

CONTENT

It is a good exercise with the definition of a limit to prove some of the assertions in this chapter. For example, the squeeze theorem can be proven by noting that since by assumption \begin{align*}\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L\end{align*} then any interval containing \begin{align*}L\end{align*} will have corresponding intervals surrounding \begin{align*}a\end{align*} such that \begin{align*}f\end{align*} and \begin{align*}h\end{align*} give values inside the interval containing \begin{align*}L\end{align*}. So if we choose the overlap (or intersection) of these intervals, then any \begin{align*}x\end{align*} in this overlap will also give values from both \begin{align*}f\end{align*} and \begin{align*}h\end{align*} inside the interval around \begin{align*}L\end{align*}. But then since by assumption \begin{align*}f(x) \le g(x) \le h(x)\end{align*} we must have that \begin{align*}g\end{align*} also gives values inside this interval.

The other results will involve a lot of notation as well and students are likely to lose the forest for the trees if teachers are not careful. The salient point of all these properties is that basically students can do exactly what they think they can do, as long as everything involved has a limit. If \begin{align*}f\end{align*} and \begin{align*}g\end{align*} have limits then \begin{align*}f+g\end{align*} has a limit and it is equal to the sum of the limits for \begin{align*}f\end{align*} and \begin{align*}g\end{align*}. Most students would be guided to suspect this and that intuition should not be shunned.

PROCESS

It is on utmost importance that students have to ability to approach limits mentally. That is, consider the problem of finding the following limit:

\begin{align*}\lim_{n \to \infty} 3 + \sqrt{\frac{1}{n}}\end{align*}

Of course, one could use the sum rule and then the constant rule and then the power rule. But it is much more powerful to simply be able to see that for large values of \begin{align*}n\end{align*}, the second term gets smaller and smaller while the first term is unchanged. So the limit is clearly \begin{align*}3\end{align*}.

The best process for learning that is accessible to all should be aimed at building this intuition. This begins with very simple limits like the one above, or other sums of basic equations. Have the students mentally approach the limiting value and think about what will happen. Perhaps this could be done as a game. Have the students put all of their materials away (or under their desks). This way they have nothing more than their minds to rely upon. Then split the class into two more or less equal teams, and have them compete to see which team can obtain a limit faster. Tell them that they are not to speak, but to raise their hand and that you will call on the first hand you see. However, once you’ve gotten a question right for your team you are not allowed to answer again until round two. Problems can be as simple as \begin{align*}\lim_{h \to \infty} \frac{1}{h}\end{align*} or \begin{align*}\lim_{h \to 0} \frac{h}{h + 1}\end{align*}. This way students will slowly be able to do these in their heads and will become more comfortable in general with limits.

The next step is to familiarize students with more complicated radicals and provide some intuition for these. For example, with a complicated limit like:

\begin{align*}\lim_{h \to \infty} \frac{4h^7 + 3h^2 - 2h + 9}{2h^7 - 6}\end{align*}

the intuition that students have just developed may falter. This is because both the numerator and the denominator seem to get large for large \begin{align*}h\end{align*} and \begin{align*}\frac{\infty}{\infty}\end{align*} is not defined. So instead they can do a little trick before employing intuition, and that is to get rid of the highest powers in \begin{align*}h\end{align*} by multiplying numerator and denominator by \begin{align*}\frac{1}{h^7}\end{align*}. That is, we note that for \begin{align*}h\end{align*} not equal to zero we have:

\begin{align*}\left (\frac{4h^7 + 3h^2 - 2h + 9}{2h^7 - 6} \right ) \left (\frac{\frac{1}{h^7}}{\frac{1}{h^7}} \right ) = \frac{4 + \frac{3}{h^5} - \frac{2}{h^6} + \frac{9}{h^7}}{2 - \frac{6}{h^7}}\end{align*}

and we can then easily use intuition on this to see that for larger and larger values of \begin{align*}h\end{align*}, all of the fractions become tiny so that:

\begin{align*}\lim_{h \to \infty} \frac{4h^7 + 3h^2 - 2h + 9}{2h^7 - 6} = \lim_{h \to \infty} \frac{4 + \frac{3}{h^5} - \frac{2}{h^6} + \frac{9}{h^7}}{2 - \frac{6}{h^7}} = \frac{4}{2} = \underline{2}\end{align*} Basically, it can be explained that limits to infinity are battles between the numerator and the denominator. If the numerator wins because it has a higher power of \begin{align*}h\end{align*}, then the fraction goes to infinity. If the denominator wins because it has a higher power of \begin{align*}h\end{align*}, then the fraction becomes tiny. And if the numerator and denominator have the same highest power, then all the other terms become tiny at big values of \begin{align*}h\end{align*} and it is only this highest power term that matters.

In order to evaluate limits as \begin{align*}h\end{align*} goes to , we can apply the same strategies by simply taking a new limit where we replace the variable \begin{align*}h\end{align*} by \begin{align*}\frac{1}{h}\end{align*} and take the limit as \begin{align*}h\end{align*} goes to infinity. Otherwise we can see that the opposite thinking works. Instead of looking at the highest powers in the numerator and the denominator, we look to the lowest powers. These are the ones which dominate as \begin{align*}h\end{align*} becomes small. This kind of intuition will be indispensible as the math becomes more advanced.

PRODUCTS

Testing this material is very straightforward, since students can simply be asked to find a series of limits with varying difficulty. However, it is a good idea to make assignments as predictive as possible. So for example, the following questions are very good ones:

1. Find the limits:

\begin{align*}& \lim_{h \to 0} \frac{(x+h) - x}{h} \\ & \lim_{h \to 0} \frac{(x+h)^2 - x ^ 2}{h} \\ & \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h} \\ & \lim_{h \to 0} \frac{(x+h)^4 - x^4}{h}\end{align*}

2. Can you find a way to calculate or guess at the following limits based upon your experience above:

\begin{align*}& \lim_{h \to 0} \frac{(x+h)^{10} - x^{10}}{h} \\ & \lim_{h \to 0} \frac{(x+h)^n - x^n}{h} \text{where}\ n \ \text{is any integer} > 0\end{align*}

## Continuity

CONTENT

The definition of continuity given in this lesson is clunky since the “conditions” are not independent. In other words, the third condition that:

\begin{align*}\lim_{x \to a} f(x) = f(a)\end{align*}

depends upon the other two being satisfied. It would be better simply to define a function as continuous at a point \begin{align*}x=a\end{align*} of its domain if the equation above is satisfied. If the limit does not exist then the equation cannot be satisfied.

This is a very intuitive definition, but the first example given in the text does not illustrate why. The reason the function \begin{align*}f(x) = \frac{(x+1)}{(x^2-1)}\end{align*} fails to be continuous at \begin{align*}x=-1\end{align*} is because \begin{align*}-1\end{align*} is not even part of the domain. That is to say, the function \begin{align*}f\end{align*} is not defined at \begin{align*}x=-1\end{align*} and therefore cannot be continuous there. This is a technicality though, and doesn’t really illustrate how the definition works.

The second example given does a much better job, and this is the content which should be used to illustrate the definition. This example uses the function:

\begin{align*}f (x) = \begin{cases} x & x \neq 1 \\ 3 & x = 1\end{cases}\end{align*}

which has the desirable property at \begin{align*}x=1\end{align*} that \begin{align*}f(x)\end{align*} approaches \begin{align*}1\end{align*} whereas \begin{align*}f(1)=3\end{align*}. This is what the definition of continuity has in mind, a jump in \begin{align*}f\end{align*} at some particular value of \begin{align*}x\end{align*}. So basically \begin{align*}f\end{align*} can fail to be continuous at a point \begin{align*}a\end{align*} because the function is not defined at \begin{align*}a\end{align*} or it jumps suddenly at \begin{align*}a\end{align*}. That’s it. The concept is not complicated, although putting the math into words requires a little thinking.   PROCESS

Some students may respond better to pictures, and some may respond better to formulas. Testing for continuity with a picture means plotting a function near the point in question. Then, if the line ends at the point in question (that is, if you would need to lift your pencil in order to draw the plot there) the function is discontinuous. This works for functions which fail to be continuous for either reason: a jump or a gap in domain. Learning from the formulas, on the other hand, means calculating limits of functions directly using rules from the earlier lesson.

A nice lesson plan would point out that occasionally we can “fix” discontinuities by simply redefining the function’s value at a single point. If this is the case, then the discontinuity is said to be removable. The piecewise function given above is a simple example since we have that:

\begin{align*}\lim_{x \to 1^+} f(x) = \lim_{x \to 1^-} f(x) = 1\end{align*}

So we can simply redefine \begin{align*}f(1)=1\end{align*} and the new function will be continuous. A similar example is the more complicated function \begin{align*}f(x)=\frac{\mathrm {Sin}(x)}{x}\end{align*}. This function is discontinuous at \begin{align*}x=0\end{align*} since it fails to be defined there, however we could make a new function from it that is continuous by setting its value at equal to the limit \begin{align*}\lim_{x \to 0} f(x) = 1\end{align*}. However, in this case the discontinuity is not said to be removable since the point was not even in the original function’s domain.

PRODUCTS

Tricky problems can be formulated to test students’ understanding of continuity. Questions which contain follow-ups asking students to describe why a function is discontinuous or continuous are good ones. Furthermore, students might be asked if a particular discontinuity is removable or can be otherwise “fixed”.

## Infinite Limits

CONTENT

The concept of infinity can be challenging to students at first. However, the concept of infinity is actually dual to the concept of zero. This is because the following are true when the corresponding limits exist:

\begin{align*}\lim_{x \to 0+} f(x) & = \lim_{x \to + \infty} f \left (\frac{1}{x} \right ) \\ \lim_{x \to 0^-} f(x) & = \lim_{x \to - \infty} f \left (\frac{1}{x} \right )\end{align*}

We have similar rules when the limits are divergent but under these circumstances the direction in which limits are taken can make things a little messy. Suffice it to say that infinity can be understood by letting the denominator of a fraction get very small.

This is actually the introduction to the topic of compactifications for the real line, which gives a nice geometric means of understanding infinite limits. The entire real line can be wrapped around a circle so that zero matches up with the very bottom of the circle and the two ends fold up on either side of the circle. We can compress these down so that they just nearly reach the top of the circle, and then call the top of the circle infinity or \begin{align*}\infty\end{align*}. Then approaching infinity just means moving along the circle towards the top and watching what happens with the target point.

PROCESS

Students will learn this concept with practice. The best way to encourage practice, though, may be through a process similar to the one we used for the section on equations. Students can be divided into groups of two or three and given a complicated function. They should try plotting the function and then examine all limits of interest including any points of discontinuity as well as \begin{align*}+ \infty\end{align*}, and \begin{align*}- \infty\end{align*}. Then they can present their function to the class with the plot and a discussion of each limit. Ideally the teacher should assure each group is progressing and try to encourage participation by each group member.

PRODUCTS

Students can demonstrate their mastery of this material by trying a number of

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