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5.1: Functions, Limits, and Continuity

Difficulty Level: At Grade Created by: CK-12
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Equations and Graphs

Much of single variable calculus centers around the graphical representation of functions. Students have been graphing functions, and working with graphs, for some years, but many will not understand that a graph is a visual representation of all solutions to an equation. If students can be brought to understand this key idea, many topics later on will become easier.

One of the first things to work on with students is the idea of substitution. For purposes of this problem, let's look at the equation \begin{align*}y=5x^5-10x^4-4x^2+8x\end{align*} Some common questions that can be asked are:

  • What are the \begin{align*}x-\end{align*}intercepts?
  • What is the \begin{align*}y-\end{align*}intercept?
  • What is the \begin{align*}y\end{align*} value when \begin{align*}x=1\end{align*}?
  • When is the \begin{align*}y\end{align*} value equal to \begin{align*}-5\end{align*}?

The easiest way to solve any of these questions is to stress that the equation is the rule that connects the two variables, and that substituting a value in for one of the variables allows the other variable to be solved. This is especially straightforward for the third question, What is the \begin{align*}y\end{align*} value when \begin{align*}x=1\end{align*}? By directly substituting \begin{align*}y=5(1)^5-10(1)^4-4(1)^2+8(1)\end{align*} and evaluating we see that \begin{align*}y=-1\end{align*}.

A little bit of translation is needed for the first two questions. Students need to understand not only what an intercept is, but also the other language used by different teachers and text books. Usually the \begin{align*}y-\end{align*}intercept is pretty standard, sometimes being referred to as the vertical intercept. The \begin{align*}x-\end{align*}intercept, however, sometimes goes by the name of horizontal intercepts, roots or zeros. The last name is probably the most useful in this case, as we are saying that a value is zero. Often students get tripped up on which variable to set to zero in these cases. The \begin{align*}y-\end{align*}intercept is found when \begin{align*}x\end{align*} is equal to zero, not the variable listed in the name. Again, solving for the \begin{align*}y-\end{align*}intercept is the easier of the two, as substituting zero in for \begin{align*}x\end{align*} yields \begin{align*}y =0\end{align*}.

Working from a given \begin{align*}y\end{align*} value adds an additional layer of complexity, as the student then needs to use additional methods to solve for \begin{align*}x\end{align*}. A tool that I will be employing frequently checking down a list of options of increasing difficulty/decreasing accuracy. This one is from my Algebra I class:

  • Can I solve directly using opposite operations? No. Method fails when variables have different exponents.
  • Can I solve using a formula? No. There is no formula for 5th degree polynomials. (not strictly true, but the formula is not one I would expect students to know)
  • Can I factor? Yes. Because it is equal to zero, and the polynomial factors, this is a valid solution method.
  • Can I use guess and check or use a computer/calculator? This always works, but is sometimes not allowed in the case of using computer help, or can be tedious and inaccurate in the case of guess and check.

So if the students can discover that after an \begin{align*}x\end{align*} is factored out of each term \begin{align*}0=x(5x^4-10x^3-4x+8)\end{align*} the fourth degree polynomial can be factored. The major clue, in this case, is that \begin{align*}5*2=10\end{align*} and \begin{align*}4*2=8\end{align*} so the factored form is: \begin{align*}0=x(5x^3-4)(x-2)\end{align*} and each factor can individually be set equal to zero and the above process repeated for each individual equation, all of which can be solved directly.

The last question is similar, but since \begin{align*}y=-5\end{align*} instead of , the polynomial can no longer be set equal to zero and then factored. This is a problem I would send straight to the graphing calculator or computer solver to get those solutions. On the graphing calculator there are two methods that work, both of which require an extra step as most calculators do not have a solver built-in. The first option is to graph the polynomial in the \begin{align*}y1\end{align*} slot and then graph the constant function \begin{align*}y=-5\end{align*} in the \begin{align*}y2\end{align*} position. After graphing, there is an intersection [INTERSECT] option under the [CALC] menu. Make sure the command is run for each point of intersection. A second method is to set the equation equal to zero and use the zero/root option under [CALC].

Relations and Functions

While most of this lesson focuses on information, the problem solving skills for finding domain restrictions will be applicable to future lessons on limits and differentiability. Let's look at a couple of functions that sometimes cause some unique issues with domain and range.

Find the domain and range of: \begin{align*}h(x)= \frac{4-x^2}{x^2-x}\end{align*}

In many cases when a question is asking you to find the domain for a given function, they are really asking For what values of \begin{align*}x\end{align*} does is this equation undefined? There are a few places to normally look in these situations:

  • Rational functions are undefined when the denominator is equal to zero
  • Even powered radicals are undefined when the inside is negative
  • Special meaning attached to problems may restrict the domain, for example “negative time” may not make sense to include.

In the case of our first problem, it is a rational function so we only need to consider the denominator and set it equal to zero: \begin{align*}0=x^2+x\end{align*} By factoring, we find \begin{align*}x =0,-1\end{align*}, so the domain is necessarily restricted by eliminating those two items. As to the range, this is a good opportunity to bring in some of the concepts about limits as we can examine the function at the numbers very close to our two undefined numbers to see that the range is infinite in both the positive and negative direction.

Find the domain and range of: \begin{align*}p(x)=\sqrt{\sin(x)}\end{align*}

A strong understanding of trig functions pays huge dividends in calculus and this is a good example. We know, from the list above, that we are looking for when \begin{align*}\sin(x)\end{align*} is negative. A student with a less than perfect grasp on the trig functions may find this difficult. A common way to find when a function will return negative values is to first find when the function is equal to zero to create intervals to test, and then test a point in each interval to see if it is negative or positive. If a student uses \begin{align*}\sin-1(0)\end{align*} to determine the intervals, they will get only a single answer, where a student with a better understanding will know that there will be more than one intersection with the \begin{align*}x-\end{align*}axis, and therefore many intervals where \begin{align*}\sin(x)\end{align*} is negative. On a single period, \begin{align*}\sin(x)\end{align*} is negative on the interval \begin{align*}(\pi,2\pi)\end{align*}. Students should also realize that \begin{align*}(3\pi,4\pi), (5\pi,6\pi), \ldots\end{align*} and \begin{align*}(-\pi,0), (-3\pi,-2\pi),\ldots\end{align*} are intervals that can't be included in the domain. Since there are infinitely many intervals, a challenge for the student is to figure out a way to write the domain. I recommend this as a short group activity to develop a plan, and then present to the class. Students will likely try to use descriptive language, which is ok, but try to steer the groups to develop a description or rule that can be written down. The usual way of expressing such a domain is: \begin{align*}D=[(2k-1)\pi,2k\pi],k \in \mathbb{Z}\end{align*} This is also a common trick for sequences that use just even, or just odd numbers, so it is worth the time to ensure students understand this notation. The range is again easy if students understand the trig functions, as \begin{align*}\sin(x)\end{align*} will reach a maximum of \begin{align*}1\end{align*}, and in this case a minimum of with the domain restrictions, and the root does nothing to change those boundaries.

Find the domain and range of: \begin{align*}r(x)=\tan(x)\end{align*}

I've included this one as it's a little bit deceiving. It doesn't appear to have any restrictions at first, but a rule that has served me well throughout calculus is to always change all trig functions to sin and cos immediately. Now it becomes \begin{align*}r(x)= \frac{\sin(x)}{\cos(x)}\end{align*} and it is clear that we should treat it as a rational function and eliminate all instances when \begin{align*}\cos(x) = 0\end{align*}. The range is infinite in each direction.

Models and Data

This is one of the finest topics to spend some time with, as much of the work done in the real world centers around modeling functions to observed data. Also, the process of selecting the correct model by finding the clues given, and the applying the correct method is an extremely valuable skill, and one that will be used frequently throughout calculus.

Knowing the general shapes of a few graphs is important for students in the future. If they have not yet, they should have memorized the general shape of:

  • Linear functions
  • Even degree polynomials (like quadratics)
  • Odd degree polynomials (like cubics)
  • Exponential functions
  • Sine

Other common graphs, such as \begin{align*}n-\mathrm{th}\end{align*} root functions, logarithmic functions and cosine are simple transformations of the graphs listed above, and do not need to be memorized explicitly on their own.

The text focuses on identifying the model from trends or graphs but there is also a way to do it analytically. The further away from the model, the harder this gets, but can often yield clues. The data needs to be arranged with the input values in order, and equally spaced. The relationships between the output values will lend clues to the type of function. The key process is taking differences between each set of output values. The following is a table with a number of functions from a single set of input values.

\begin{align*}& x && f(x) && g(x) && h(x) && q(x) && r(x) \\ & -3 && -9 && 10 && -26 && .0156 && 1 \\ & -2 && -7 && 5 && -7 && .0625 && 0 \\ & -1 && -5 && 2 && 0 && .25 && -1 \\ & 0 && -3 && 1 && 1 && 1 && 0 \\ & 1 && -1 && 2 && 2 && 4 && 1\\ & 2 && 1 && 5 && 9 && 16 && 0 \\ & 3 && 3 && 10 && 28 && 64 && -1 \\ & 4 && 5 && 17 && 65 && 256 && 0 \\ & 5 && 7 && 26 && 126 && 1024 && 1\end{align*}

If you take the each output value for function \begin{align*}f(x)\end{align*} and subtract the one previous, you get a constant answer,\begin{align*}2\end{align*}. If all these “first differences” are equal, then the function is a linear function.

For function \begin{align*}g(x)\end{align*} the sequence of differences are: \begin{align*}-5,-3,-1,1,3,5,7\end{align*}. The next step is to look at the differences of this sequence, which are all equal to \begin{align*}2\end{align*}. If the “second differences” are equal, then the function is a quadratic.

Start \begin{align*}h(x)\end{align*} the same way, finding the first sequence of differences to be: \begin{align*}19,7,1,1,7,19,37,61\end{align*}. The second sequence of differences is: \begin{align*}-12,-6,0,6,12,18,24\end{align*} which makes the “third differences” all equal at \begin{align*}6\end{align*}. This is a cubic function, and the pattern holds for all higher degree polynomials.

No sequence of differences will ever start getting close to being equal, so we can rule this out as a polynomial. The next technique to attempt is to inspect the ratios of the outputs. In this case, if we divide each entry by its previous, all the ratios equal \begin{align*}4\end{align*}. If the ratios are equal, the function is an exponential function.

The toughest is the trig functions, which is what \begin{align*}r(x)\end{align*} is. Sometimes you can only determine it by process of elimination, or have enough entries to identify that the outputs are periodic, such as \begin{align*}r(x)\end{align*} in this case.

Once a model is selected a set of \begin{align*}x-y\end{align*} pairs are chosen to solve for missing coefficients as a system of equations. As many pairs are needed as missing elements. For example, to find the equation for \begin{align*}h(x)\end{align*}, we might set up a system such as:

\begin{align*}-7 & = a(-2)^3+b(-2)^2+c(-2)+d \\ 0 & =a(-1)^3+b(-1)^2+c(-1)+d \\ 1 & =a(0)^3+b(0)^2+c(0)+d \\ 2 & =a(1)^3+b(1)^2+c(1)+d \end{align*}

From here there is the option to use matrices, or elimination/substitution to find the coefficients.

Sometimes observed data is not going to yield exact answers, so a best approximation will need to be made. Working with a few problems with exact outputs will help to give the experience to sense what is the correct model choice.

The Calculus

In this conceptual treatment of calculus centers around the understanding of small approximations all adding up to an exact answer. As a conceptual lesson, there will not be any specific problems to solve here, but many can be found in later sections.

A challenge of teaching calculus is where to start. Do you try to make sure students have a conceptual foundation for what they are doing computationally later, or do you dive in into the computation and then fill in the meaning of those computations later? Either method has it's faults, as there will need to be some “hand-waving” as some ideas and techniques will not be filled in until later. Calculus is does two things for the understanding. First, students begin to understand why the curriculum in Algebra-Geometry-Math Analysis is structured the way it is: for the application to calculus problems. Second, Calculus makes much more sense after the entire course is completed. Therefore students will need encouragement and support over the next lessons which involve many tricky and long problems that have the potential to frustrate students. Avoid creating a mutiny by giving them the confidence to “fight” through it for now, as things will start to come together as time goes on.


A nice way to guide students to understand the nature of limits, as well as introduce some of the important concepts of calculus is to look at the concept of instantaneous and average velocity. Students often have an understanding of each of those concepts separately; their experience of the speedometer in the car, or the radar gun readings for instantaneous velocity, where most of the problems they have done in math classes all relate to average velocity.

Problem: A cyclist's position in a \begin{align*}1 \;\mathrm{kilometer}\end{align*} time trial can be modeled by the equation \begin{align*}s(t)=\frac{-t^3}{300}+ \frac{1}{2} t^2\end{align*} where \begin{align*}s(t\end{align*}) is the meters traveled in time \begin{align*}t\end{align*} in seconds. What is the rider's average speed? What is the rider's speed when they cross the finish line? What is the rider's speed at \begin{align*}50 \;\mathrm{seconds}\end{align*}?

For the solution, the problem needs a little bit of working. Students should be familiar with the fact that the rate is the change in distance over time. A quick note on that. There are two things worth stressing at this point. A strong tool to use in both math and science classes is to gather what information of formulae you need to access through the units of the answer. In this case, speed is given in m/s, so distance and time are needed. The next is to start understanding the relationship between slope and rates. The rise-over-run mantra should be followed with “What is the meaning of the rise, and the meaning of the run, in this case?” For this problem, the vertical axis is position, the horizontal time, so the change in position over the change in time gives the slope, which is the speed in this case.

Since the students are looking for the rate, and know that they need the distance and the time. The distance is given, \begin{align*}1\mathrm {km}\end{align*} or \begin{align*}1000m\end{align*}, but the time is not. Students will need to find the time it takes to cover that distance, but ideally, should not need to be told explicitly this is what they need. Individuals or groups should try to work to discover this on their own. To find the time, they should go through the checklist to see if they can solve the equation directly, but with minor exception, polynomials of degree \begin{align*}3\end{align*} and higher will most easily be solved by graphing and finding points of intersection, which yields that it takes \begin{align*}56.7 \;\mathrm{seconds}\end{align*} to travel this distance. The average speed then being \begin{align*}17.6 \;\mathrm{m/s}\end{align*}.

Now for the more important question, which is about the instantaneous speed. As groups, think-pair-share, or as a class discussion students should be asked to contribute their ideas as to how to find the answer. Some hints can be given about relating slope to rate, and what the slope at that point would be. Gropus may also come up with the idea that the change in time for instantaneous velocity is , which can't be used, due to division by zero, but this is a valuable observation. Much of calculus is about very good approximations, so ask students what a better approximation of the instantaneous velocity might be. After getting contributions, students should begin to see that choosing points that are very close, infinitely close, together will give the closest answer.

This is a good motivation as to why limits are important. It is valuable in math to look at very close approximations, and if you are close enough, it is as good, and accepted as, the exact answer.

Evaluating Limits

The order that the different techniques are presented in is also the algorithm for solving limits. To put it all in one place:

  • Direct substitution. Always try to simply put the number into the expression
  • Factor and divide.
  • Separate and simplify using properties of limits.
  • Apply special known limits.
  • Use an analytical technique, such as the squeeze theorem, or l'Hopitals rule.

Some special known limits include:

\begin{align*}\lim_{x \to 0} \frac{\sin(x)}{x} = 1 \lim_{x \to 0} \frac{1 - \cos(x)}{x} = 0 \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2}\end{align*}

Knowing these are helpful, especially as there are often a couple of problems on the AP examination that are much easier if you know them.

A couple of tough examples:

\begin{align*}\lim_{x \to 27} \frac{x - 27}{x^{\frac{1}{3}} - 3}\end{align*}

Always try to plug in the number, which predictably does not work in this case. Now it's time to factor, or use other algebraic methods. Typically, when roots are involved, multiplying by the conjugate is the first step, in this case multiplying by \begin{align*}x + 27\end{align*} does not get closer to a simpler expression. The key here, as with all factoring problems, is to try to find the relationship between numerator and denominator that will allow for the elimination of a factor. In this case, I notice that each term in the denominator cubed separately becomes the corresponding term in the numerator, so I will look to factor the numerator as a cubic. Remembering the form for a difference of cubes: \begin{align*}a^3-b^3=(a-b)(a^2+ab+b^2)\end{align*} and thinking of \begin{align*}x-27=\left (x^{\frac{1}{3}} \right )^3-3^3\end{align*} results in a factoring and elimination as follows:

\begin{align*}\lim_{x \to 27} \frac{\left (x^{\frac{1}{3}} - 3 \right ) \left (x^{\frac{2}{3}} + 3x^{\frac{1}{3}} + 9 \right )}{x^{\frac{1}{3}} - 3} = \lim_{x \to 27} x^{\frac{2}{3}} + 3x^{\frac{1}{3}} + 9\end{align*}

Which can be evaluated directly as equal to \begin{align*}27\end{align*}.

\begin{align*}\lim_{x \to 0} \frac{\sin(5x)}{3x}\end{align*}

Again, always at least try to plug the number directly in. Further, this is not factorable in any useful way, although a brief glance as double and half angle rules are useful from time to time. There is no useful way to separate these out using the limit properties, but it should be clear that the next step, using a way to relate to special known limits, is helpful. This one looks very close to \begin{align*}\lim_{x \to 0} \frac{\sin(x)}{x}=1\end{align*} and only needs the following algebraic manipulation:

\begin{align*}\lim_{x \to 0}\frac{\sin(5x)}{3x} \times \frac{5}{5} = \lim_{x \to 0} \frac{\sin(5x)}{5x} \times \frac{5}{3}\end{align*}

To be completely rigorous, a change of variable should be used here, such as \begin{align*}5x = u\end{align*} but it doesn't change the problem and most solutions will omit this step, seeing simply that the limit of the first factor is equal to one, and that the answer is \begin{align*}5\end{align*} over \begin{align*}3\end{align*}.


Continuity is a sticky subject. A concept that is easy to grasp, but putting a rigorous analytical definition to is trouble. One only needs to look at how the accepted definition changes from the first time it appears, through analysis and then to advanced topics like measure theory. It is a useful exercise to ask students to try to come up with a solid definition of their own.

An important result from continuity is the Intermediate Value Theorem. Students are not often asked to apply theorems in proofs at this level, but the skill is valuable. Also, there are sometimes free-response questions on the AP examination that ask for verification of existence, which is really asking for a slightly lighter version of proof.

The classical application of the Intermediate Value Theorem is the question:

Show that all polynomials of degree \begin{align*}5\end{align*} have at least one real root. (the more general question involving polynomials of odd degree is proven the same way, but introduces some difficulty for students in working with the general form of polynomials. I would not want to risk confusing students with variable coefficients, subscripts and missing terms in the middle, especially as it is not essential for the problem at this time.)

This is a nice introduction to analytical proofs. Here is the process for figuring out and writing this proof:

  • Is there a theorem that may be applied?

In this case, yes, and you should probably explicitly state that the Intermediate Value Theorem should be used.

  • How can you meet the conditions between “If” and “Then”?

A difficult part for students in writing what I call “grown-up” proofs is that they feel lost. The question alone is a little too open-ended to always know how to proceed, so grabbing onto necessary conditions for a theorem is a great way to start working with the problem, even if sometimes it doesn't work out in the end. In this case, we need continuity and we will need to show that there is an interval where \begin{align*}f(a)<0<f(b)\end{align*}.

  • Are there additional theorems, or pieces of information, needed to get the needed conclusion?

In this case, no. The intermediate value theorem is about all that is needed. A little bit of work with limits may also be used.

Proof: With the \begin{align*}5\mathrm{th}\end{align*} degree polynomial, \begin{align*}p(x)=ax^5+bx^4+cx^3+dx^2+ex+f\end{align*} we can take a look at two limits: \begin{align*}\lim_{x \to - \infty} p(x)\end{align*} and \begin{align*}\lim_{x \to \infty}p(x)\end{align*}. Since the limits are to infinity, the first term will dominate the others and we only need to concern ourselves with the sign of that term. Case \begin{align*}1 : a\end{align*} is positive. This would result in \begin{align*}\lim_{x \to \infty}p(x)=- \infty\end{align*} and \begin{align*}\lim_{x \to \infty}p(x)= \infty\end{align*} due to the odd exponent. Since \begin{align*}- \infty < 0 < \infty\end{align*} the intermediate value theorem states that the polynomial must be equal to zero at some point in the reals. Case \begin{align*}2: a\end{align*} is negative. In this case the signs of the above limits both switch, which has no effect on the inequality and the intermediate value theorem still holds.

Note: I did not establish the fact that all polynomials are continuous. This is true, and is given as fact in many texts. It is not valuable to stress this point as accessing the other parts of the intermediate value theorem are more important.

Infinite Limits

A place where students can become confused here is with the difference between an indeterminate form of a limit and one that has real meaning. This is especially difficult as infinity is a concept, not a number, but seem sometimes like it is treated as a number. An example: \begin{align*}lim_{x \to \infty} 2x^4 - 35x^3\end{align*} Which, if we substitute we get: \begin{align*}\infty - \infty\end{align*} which is an indeterminate form. By factoring out an \begin{align*}x^3\end{align*} we get \begin{align*}lim_{x \to \infty} x^3 (2x - 35) = \infty\end{align*} which is not an indeterminate form. The difference being that while students will be tempted to say that \begin{align*}\infty - \infty = 0\end{align*} no such assumption can be made. However, there is no circumstance in which \begin{align*}\infty \times \infty\end{align*} does not go to infinity. Some of the subtleties can cause confusion.

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