A.4 Lab 2.1: Find a Limit Using Approximations

Calculus Home Limits & Continuity A. The Concept of Limits A.4 Lab 2.1: Find a Limit Using Approximations

On this screen we’re going to determine the limit of a function at a point where the function is undefined, using an approach that’s very similar to that of Lab 1.1 at the end of Chapter 1.

Reminder: \[\text{The limit of the function $f(x)$ as $x$ approaches $a$, written} \] \[\lim_{x \to a} f(x) = L \] \[\text{is a number $L$ (if one exists)}\] \[\text{such that $f(x)$ is $\underbrace{\text{as close to}}_{\Large{\text{within } \pm
\epsilon} \text{ of}}$ $L$ as we want whenever $x$ is $\underbrace{\text{sufficiently close to}}_{\Large{\text{within }
\pm \delta \text{ of}}}$ $a.$ }\]

The function we’re going to consider for this lab is $g(x) = \dfrac{2^x\, -\, 2}{x-1}.$ As you can see, this function is undefined for $x=1$ since the denominator would be zero there. A graph of the function thus has a hole at $x=1,$ as shown.
Graph of y=g(x), with a hole at x=1. The unknown y-value of that hole is labeled L.

Graph of y=g(x), with a hole at x=1. The unknown y-value of that hole is labeled L.

We don’t know (or pretend to not know, anyway) the exact height, the y-value, of that hole. But we have marked its approximate location on the y-axis with the label “L” since the (currently unknown) exact value of this y-location is
\[\lim_{x \to 1}\overbrace{\frac{2^x\, -\, 2}{x-1}}^{g(x)} = L\] We’re going to work to determine the value of L to whatever level of precision we would like. While this may seem like a silly exercise, we’re actually engaging in a deep use of Calculus tools (as well as some basic ideas of data analysis).

Note: If you have taken Calculus before, you may know a fancy way to compute this limit exactly. You could also use any number of online resources to find its value easily. Please don’t. Later in the course we’re going to need the crucial ideas and tools we’re developing here, and so ask that you complete the activities below. In the end we don’t really care what “the answer” is to this particular limit; instead, we care that you’re working to develop an understanding of Calculus and its tools that will allow you to do whatever you want in the future.

Lab Activity: Approximate $\displaystyle{L = \lim_{x \to 1}\frac{2^x – 2}{x-1}}$

This lab is based on work done by the CLEAR Calculus Project at Oklahoma State University.
From the Project’s website: “Project CLEAR Calculus is a research-based effort to make calculus conceptually accessible to more students while simultaneously increasing the coherence, rigor, and applicability of the content learned in the courses.”
We at Matheno appreciate the work CLEAR Calculus has done to help students learn Calculus better, and are happy to build off of their efforts.

As we did in the earlier lab, let’s use our Questions for Approximations to frame our work. We’ll consider the first of the questions now. Please answer each Question for yourself before reading our discussion:
Questions for Approximations [Reference]

1. What are you approximating?

Open to view answer

We are approximating $\displaystyle{L = \lim_{x \to 1}\frac{2^x\, -\, 2}{x-1}}$. Visually, this is the “height of the hole” in the graph above.

Close answer

2. How will you generate an overestimate for the value we’re after? An underestimate?
[Hint: Look at the graph, and think about where you would “sample” the function’s data to produce a value that is explicitly larger than or less than the height of the hole.]

Open to view answer

To approximate the value L we compute the output of the function $g(x) = \dfrac{2^x\, -\, 2}{x-1}$ for values of x that close to, but not equal to, $x=1.$ From the graph we can see that the function increases as x increases. Hence to generate an overestimate we use a value of $x \gt 1,$ and to generate an underestimates we use a value of $x \lt 1.$

Close answer

We’ll discuss the remaining questions in a bit. As a reminder, they are:

3. WThe difference between your current overestimate and underestimate values is your “error bound.” What is its current value?
4. How can the error bound be made smaller than any predetermined value?

For now, let’s get to approximating!

Part I: Upper Bound, Estimate from the Right

The interactive Desmos graph below shows the graph of our function of interest, $g(x)=\dfrac{2^x – 2}{x-1}.$ As we start to estimate the limit of the function at $x=1,$ let’s first develop an upper bound by finding an overestimate using a point with x-value greater than $x=1.$ You can move the point on the curve to wherever you’d like, as long as $x_1 > 1.$

Graph of $g(x) = \dfrac{2^x – 2}{x-1}$ versus x

Your current upper bound for the value of L is
L ≤ g(1.76) = 1.8249753

When you’re ready, click the button to add this first value to your data table. It doesn’t matter where exactly the point is, or how close it is to $x=1$; we just need a first value to get started here.

Table I: Data Collection for $\displaystyle{L = \lim_{x \to 1}\frac{2^x – 2}{x-1}}$, with $x_1 \gt 1$ (Overestimate)
Data Point	$x_1$	$d = x_1 – 1$	L Upper Bound Value: $g(x_1)$

Great! Since this first value (with $x_1 > 1$) is an overestimate of L, we have thus far determined that
\[ L \le 1.8249753|\]
Notice that we have added the horizontal distance d = to the table, which is the distance from $x=1$ to your point $x_1$: $x_1 = 1+d$. This horizontal distance is also now visible on the graph above so you can see it. (You may need to zoom in for it to be visible.)

You’ll add more data as we continue. Please proceed to Part II.

Part II: Lower Bound, Estimate from the Left

Rather than having you place a point with x-value $x_2 \le 1,$ we’ve instead used the value of d you set in Part I to be an equal distance to the left of $x = 1$: $x_2 = 1 – d.$

Specifically, you set $d = \text{[you have not yet clicked the “Add data” button in Part I]}$ and so

$x_2 =\, $ ?

and we see in the graph below that your current lower bound for the value of L is

$g(x_2) = $[You have not yet clicked the “Add data” button in Part I] $\le L$

Graph of $g(x) = \dfrac{2^x – 2}{x-1}$ versus x

Once you’ve verified that the point $x_2 = 1 – d,$ where d is the value you chose in Part I, please click the “Add data to the table” button below.

Table I (modified): Data Collection for $\displaystyle{L = \lim_{x \to 1}\frac{2^x – 2}{x-1}}$
Data Points	$d$	$x_2 = 1 – d$	L Lower Bound Value: $g(x_2)$	$x_1 = 1 + d$	L Upper Bound Value: $g(x_1)$

Hence from your choice of d above, so far we know that the value of L lies in the range
\[\text{Bounds not yet set}\]
With the first overestimate and underestimate in place, we can address our third Approximation Question:

3. The difference between your current overestimate and underestimate values is your “error bound.” What is its current value?

The current size of your error bound, the difference between the upper and lower bounds, is
\[\]
Our fourth and final Approximation Question is

4. How can the error be made smaller than any predetermined bound?

You probably have an answer to this question in mind. You’ll get to implement it in Part III below.

Part III: Decrease the error bound to within a predetermined size

As happened on the preceding screens, it can become tough to see what’s going on with both the function’s output and input as we zoom in more and more. Hence, as we did earlier, we’re going to split the graph into two linked graphs, the top one focusing on the function’s output and the lower one focusing on the function’s input.

We’ve set the initial size of d set to the value we used in Parts I and II above. You can see visually the values for $g(x_1)$ and $g(x_2)$ bounding the value of L.

Below the graphs is a slider that allows you to set the value of d. Give it a try to see how it works.

Graph of $g(x) = \dfrac{2^x – 2}{x-1}$ versus x: Focus on Function Output

Graph of $g(x) = \dfrac{2^x – 2}{x-1}$ versus x: Focus on Function Input

d = 0.48

Table I (further modified): Data Collection for $\displaystyle{L = \lim_{x \to 1}\frac{2^x – 2}{x-1}}$
Data Points	$d$	$x_2 = 1 – d$	L Lower Bound Value: $g(x_2)$	$x_1 = 1 + d$	L Upper Bound Value: $g(x_1)$	Error bound: $g(x_2) – g(x_1)$

The first row of the data table immediate above has your values from Parts I and II. As we wrote at the end of Part II, from those data we know that
\[\text{Bounds not yet set}\]
and the current error bound is
\[ \]
We’ve now used red text for those values since we are going to work to decrease the error bound and thereby improve our estimate to within a particular tolerance that you have not yet achieved.
Specifically, we are setting your target tolerance level as

target error bound < ?

Please adjust the size of d and generate another set of data. You will then see whether your new range for the value of L is sufficiently small, or if you need to make d smaller still. You can make these adjustments as many times as you’d like.

Part IV: Decrease the error bound even more

To complete the lab, let’s make our estimate of L better still.

In particular, we are now setting your target tolerance level to be
error bound < 0.001. That means you will determine the value of L to within ± 0.001, three decimal places.

Graph of $g(x) = \dfrac{2^x – 2}{x-1}$ versus x: Focus on Function Output

Reminder: To change only the graph’s vertical scale on desktop or laptop, place your cursor near the left edge of the calculator and hold the SHIFT key while you scroll.

Graph of $g(x) = \dfrac{2^x – 2}{x-1}$ versus x: Focus on Function Input

d = 0.01

Table II: Further Data Collection for $\displaystyle{L = \lim_{x \to 1}\frac{2^x – 2}{x-1}}$
Data Points	$d$	$x_2 = 1 – d$	L Lower Bound Value: $g(x_2)$	$x_1 = 1 + d$	L Upper Bound Value: $g(x_1)$	Error bound: $g(x_2) – g(x_1)$

And as one final question . . .

Now that you have found a value of d that gives an error bound < 0.001, what is the largest value of d to within four decimal places that meets this criterion?

The largest value of d that still gives an error bound of 0.001 is:

0.0007 0.0010 0.00111

[End of lab]

We think it is important that you have the experience, once, of “zeroing in” on a limit by generating the function’s output values for input values of x that are increasingly close to $x=a,$ in order to gain an even better understanding of what “the limit” really represents. While we won’t ask you to repeat the detail of this process for other functions, we hope that you will keep the Approximation Framework in mind and imagine applying it to the functions on the next screen — where we will look at some situations where the limit does, and does not, exist.

For now, if you have any questions or comments about anything on this screen, or about limits in general, please do our Learning Community a favor and post them on the Forum!

The Upshot

We can estimate the limit of a function to within any error bound we choose by sampling the function’s output at input values sufficiently close to the point of interest.

Reference for Oehrtman’s Five Questions:
Oehrtman, M. (2008). Layers of abstraction: Theory and design for the instruction of limit concepts. In M. P. Carlson & C. Rasmussen (Eds.), Making the Connection: Research and Teaching in Undergraduate Mathematics Education, (MAA Notes, Vol. 73, pp. 65-80). Washington, DC: Mathematical Association of America.

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