Estimating Limit Values from Tables (AP Calculus AB 1.4)

Sometimes all you have is numbers: a few values from an experiment, a table printed in a question, or a function whose formula you can't yet simplify. A limit only asks where the outputs head as the input closes in on a point, so a table of outputs at inputs closing in on that point lets you estimate it.

Reading a limit from a table

The function $g(x) = \dfrac{x^3 - 8}{x - 2}$ is undefined at $x = 2$ . Here are its values at inputs approaching 2 from each side:

$x$	$1.9$	$1.99$	$1.999$	$2$	$2.001$	$2.01$	$2.1$
$g(x)$	$11.41$	$11.9401$	$11.994001$	undefined	$12.006001$	$12.0601$	$12.61$

Read it from the outside in. Moving right along the left half, the outputs rise toward 12; moving left along the right half, they fall toward 12. Both sides head to the same number, so the limit appears to be 12:

\lim_{x \to 2} \frac{x^3 - 8}{x - 2} \approx 12.

The pattern carries more information than the last entry. Each time the input's distance from 2 shrinks by a factor of 10, the output's distance from 12 shrinks by about a factor of 10 too ( $0.59$ , then $0.0599$ , then $0.005999$ on the left). Steady shrinking like that is what a real limit looks like in a table. Here the algebra confirms it: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ , so for $x \neq 2$ , $g(x) = x^2 + 2x + 4$ , which is near $4 + 4 + 4 = 12$ when $x$ is near 2. Topic 1.6 develops that kind of algebra; the table got there without it, but only as an estimate.

What a table can and cannot show

A table lists finitely many inputs, and the limit is about all inputs near $c$ . So the table supports an estimate, never a proof: the function could do something different at inputs nobody listed. Within that limit, each pattern in a table points to a conclusion.

How many digits to claim: trust the digits that the closest values on both sides agree on. When the values on one side rise and those on the other side fall, the limit is trapped between the two closest values, and only the digits shared by everything in that range are reliable. Example 2 below works through this.

Choosing your own inputs

When you build the table yourself on a calculator, you choose the inputs, and two choices can wreck the estimate.

Inputs that fall into a pattern. Take $s(x) = \sin\left(\dfrac{\pi}{x}\right)$ near 0, with the natural-looking inputs $x = \pm 0.1, \pm 0.01, \pm 0.001$ . Then $\frac{\pi}{x}$ is $\pm 10\pi$ , $\pm 100\pi$ , $\pm 1000\pi$ , all whole multiples of $\pi$ , and the sine of every one of them is 0. The table is a column of zeros and suggests the limit is 0. But at $x = 0.4, \frac{2}{9}, \frac{2}{101}, \dots$ , inputs that also crowd in on 0, the angle is $2.5\pi, 4.5\pi, 50.5\pi, \dots$ and $s(x) = 1$ . The function oscillates between $-1$ and 1 near 0 (the oscillation of topic 1.3), so the limit does not exist. The tidy inputs happened to land only where the oscillation crosses 0. Use some untidy inputs too, like $0.0137$ or $-0.0049$ , and the disagreement shows up at once.

Inputs that are too close. A calculator keeps a fixed number of significant digits, and subtracting two nearly equal numbers loses most of them. Here is $\dfrac{1 - \cos x}{x^2}$ , whose limit at 0 is $\frac{1}{2}$ , computed in the standard double precision of a computer (it is an even function, so the negative inputs give the same values):

$x$	$0.1$	$0.01$	$0.001$	$10^{-6}$	$10^{-7}$	$10^{-8}$
$\frac{1 - \cos x}{x^2}$	$0.49958$	$0.4999958$	$0.49999996$	$0.50004$	$0.4996$	$0$

The values settle beautifully toward $0.5$ , then wander, then collapse to 0. At $x = 10^{-8}$ , $\cos x$ is so close to 1 that the computer stores it as exactly 1, and the numerator becomes $1 - 1 = 0$ . The 0 is round-off error, not mathematics. A calculator breaks down in the same way, at inputs that depend on how many digits it carries. Distances of $0.1$ , $0.01$ , $0.001$ , and maybe $0.0001$ are usually enough; if the values stop settling as you go closer, trust the earlier, steadier ones.

Worked examples

Worked exampleExample 1: building a table (calculator)

Use a table of values to estimate $\lim\limits_{x \to 1} \dfrac{\ln x}{x - 1}$ .

Show solution

The function is $\frac{0}{0}$ at $x = 1$ , so compute values on both sides at distances $0.1$ , $0.01$ and $0.001$ :

$x$	$0.9$	$0.99$	$0.999$	$1.001$	$1.01$	$1.1$
$\frac{\ln x}{x - 1}$	$1.05361$	$1.00503$	$1.00050$	$0.99950$	$0.99503$	$0.95310$

From the left the values fall toward 1, from the right they rise toward 1, and the distance from 1 shrinks by about a factor of 10 at each step. The limit appears to be 1. Both sides were needed: the left values alone overshoot, and the right values alone undershoot, so only together do they trap the answer between $0.99950$ and $1.00050$ .

Worked exampleExample 2: how many digits to trust (calculator)

Estimate $\lim\limits_{x \to 0} (1 + x)^{1/x}$ to three decimal places.

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Compute values at small inputs on both sides:

$x$	$-0.001$	$-0.0001$	$0.0001$	$0.001$
$(1 + x)^{1/x}$	$2.719642$	$2.718418$	$2.718146$	$2.716924$

The values from the left decrease and those from the right increase, so the limit lies between the two closest values, $2.718146$ and $2.718418$ . Every number in that range rounds to $2.718$ , so the limit is $2.718$ to three decimal places. A fourth decimal is not justified: the range includes numbers rounding to both $2.7181$ and $2.7184$ . (The exact limit is the number $e \approx 2.71828$ , which is one way $e$ can be defined.) Going much closer to 0 is not the fix for more digits, because of the round-off problem above.

Worked exampleExample 3: a table with no formula (no calculator)

The functions $h$ and $m$ are defined for all $x \neq 1$ , and $h(1) = 3$ . The table below gives selected values.

\begin{array}{c|ccc|c|ccc} x & 0.9 & 0.99 & 0.999 & 1 & 1.001 & 1.01 & 1.1 \\ \hline h(x) & 2.61 & 2.9601 & 2.996001 & 3 & -0.996 & -0.96 & -0.6 \\ m(x) & 100 & 10{,}000 & 1{,}000{,}000 & \text{undefined} & 1{,}000{,}000 & 10{,}000 & 100 \end{array}

(a) Estimate $\lim\limits_{x \to 1^-} h(x)$ and $\lim\limits_{x \to 1^+} h(x)$ . Based on the table, does $\lim\limits_{x \to 1} h(x)$ exist?

(b) A classmate says that $\lim\limits_{x \to 1} h(x) = 3$ because $h(1) = 3$ . Respond to the claim.

(d) Can the table prove your answer to part (a)? Explain.

Show solution

(a) From the left, $2.61$ , $2.9601$ , $2.996001$ approach 3, so $\lim\limits_{x \to 1^-} h(x)$ appears to be 3. From the right, $-0.6$ , $-0.96$ , $-0.996$ approach $-1$ , so $\lim\limits_{x \to 1^+} h(x)$ appears to be $-1$ . The one-sided limits appear to differ, so the limit appears not to exist.

(b) The value $h(1) = 3$ plays no part in the limit, which depends only on inputs near 1 and not equal to 1. It happens to match the left-hand limit, but the right side heads to $-1$ , so the two-sided limit appears not to exist whatever $h(1)$ is.

(c) The values grow without bound on both sides: each time $x$ gets 10 times closer to 1, $m(x)$ gets 100 times larger. So $m(x)$ appears to increase without bound as $x$ approaches 1, that is, $\lim\limits_{x \to 1} m(x) = \infty$ (the limit does not exist as a real number), and the graph of $m$ appears to have a vertical asymptote at $x = 1$ .

(d) No. The table shows six inputs near 1, and the limits depend on all inputs near 1. The function could, for example, jump to some other value at $x = 0.9999$ without the table showing it. The table supports the estimates; a formula, a graph known to be exact, or more information about $h$ would be needed to prove them.

Where students lose points

Check yourself

Question 1
For a function $f$ , $f(1.9) = 3.8$ , $f(1.99) = 3.98$ , $f(1.999) = 3.998$ , $f(2) = 7$ , $f(2.001) = 4.002$ , $f(2.01) = 4.02$ and $f(2.1) = 4.2$ . Based on these values, what is the best estimate of $\lim\limits_{x \to 2} f(x)$ ?
1. A
  $7$
2. B
  $4$
3. C
  $3.998$
4. D
  The limit does not exist.
Show answer
B
The outputs on both sides close in on 4, and the limit ignores the value at $x = 2$ . Choice A is $f(2)$ ; choice C is a single output from one side, still carrying the error of an input $0.001$ away from 2.
Question 2
A table gives only $g(3.1) = 8.2$ , $g(3.01) = 8.02$ and $g(3.001) = 8.002$ . What can you conclude about limits of $g$ at $x = 3$ ?
Show answer
The right-hand limit $\lim\limits_{x \to 3^+} g(x)$ appears to be 8. Nothing can be concluded about the left-hand limit or the two-sided limit.
Every input in the table is greater than 3, so the table describes only the approach from the right. The left side could head somewhere else entirely, so the two-sided limit is unknown.
Question 3
Using a calculator in radian mode, $\dfrac{\sin(3x)}{x}$ equals about $2.95520$ at $x = \pm 0.1$ , $2.99955$ at $x = \pm 0.01$ , and $2.9999955$ at $x = \pm 0.001$ . Estimate $\lim\limits_{x \to 0} \dfrac{\sin(3x)}{x}$ .
Show answer
$3$
The function is even, so both sides give the same values, and they close in on 3, with the distance from 3 shrinking by a factor of about 100 each time $x$ shrinks by 10. The limit appears to be 3.
Question 4
To estimate $\lim\limits_{x \to 0} f(x)$ with a calculator, which set of inputs is best?
1. A
  $0.1, 0.01, 0.001$
2. B
  $\pm 0.1, \pm 0.01, \pm 0.001$
3. C
  $\pm 1, \pm 2, \pm 3$
4. D
  $\pm 10^{-12}, \pm 10^{-14}, \pm 10^{-16}$
Show answer
B
A two-sided limit needs inputs on both sides of 0 that get close to 0. Choice A checks only the right side; choice C never gets close to 0; choice D is so close that round-off in the calculator can swamp the true values.

Go deeper

OpenStax, Calculus Volume 1, 2.2 The Limit Of A Function