Math Core

Before you learn algebraic techniques for evaluating limits, you need to be able to see a limit. On the AP exam, limits are often given through a graph or a table of values rather than a formula, and you are expected to read them off accurately and to recognize when a limit does not exist.

Estimating limits from a table

A table lets you watch the outputs as the inputs close in on $c$ from both sides. Take

f(x) = \frac{\sin x}{x}

(with $x$ in radians) near $x = 0$ , where $f$ is undefined.

$x$	$-0.1$	$-0.01$	$-0.001$	$0.001$	$0.01$	$0.1$
$f(x)$	$0.998334$	$0.999983$	$1.000000$	$1.000000$	$0.999983$	$0.998334$

(The entries at $\pm 0.001$ are rounded; the true values are slightly less than $1$ .) From the left and from the right the outputs close in on $1$ , so the table suggests

\lim_{x \to 0} \frac{\sin x}{x} = 1.

You will prove this is exactly right in the lesson on the squeeze theorem.

A good table has three features:

Inputs on both sides of $c$ , so you can check that the left-hand and right-hand behavior agree.
Inputs that get progressively closer to $c$ , such as distances of $0.1$ , $0.01$ , $0.001$ .
A clear trend in the outputs, not just one or two values.

Worked example: Estimating with a table

Use a table to estimate $\displaystyle \lim_{x \to 0} \frac{e^x - 1}{x}$ .

Solution. Evaluate on both sides of $0$ (values rounded to six decimal places):

$x$	$-0.1$	$-0.01$	$-0.001$	$0.001$	$0.01$	$0.1$
$\dfrac{e^x - 1}{x}$	$0.951626$	$0.995017$	$0.999500$	$1.000500$	$1.005017$	$1.051709$

The left-hand values increase toward $1$ and the right-hand values decrease toward $1$ . The limit appears to be $1$ .

Common mistake

A table can only suggest a limit; it can never prove one. Tables can even be badly misleading. For $g(x) = \sin\left(\dfrac{\pi}{x}\right)$ , the inputs $x = 0.1, 0.01, 0.001$ all give $g(x) = \sin(10\pi), \sin(100\pi), \sin(1000\pi)$ , which all equal $0$ . Yet $g$ oscillates between $-1$ and $1$ infinitely often near $0$ , and its limit does not exist. A handful of carefully chosen inputs happened to hide the oscillation.

Estimating limits from a graph

On a graph, a limit is a statement about heights. To find $\displaystyle \lim_{x \to c^-} f(x)$ , trace along the curve from the left toward the vertical line $x = c$ and ask what height you are approaching. Do the same from the right for $\displaystyle \lim_{x \to c^+} f(x)$ .

Two graphing conventions matter a great deal:

An open circle marks a point that is not on the graph (a hole or an excluded endpoint).
A filled dot marks a point that is on the graph; it shows the function value.

Reading a limit at x = c from a graph

Find the height the graph approaches from the left: that is $\displaystyle \lim_{x \to c^-} f(x)$ .
Find the height the graph approaches from the right: that is $\displaystyle \lim_{x \to c^+} f(x)$ .
If the two heights agree, the two-sided limit is that common height. If not, the two-sided limit does not exist.
Only then look for a filled dot on the line $x = c$ : that is $f(c)$ , which is a separate question.

Worked example: Reading limits from a graph

The graph of $F$ is shown. Find each value, or state that it does not exist.

The graph of F. Filled dots at (1, 3) and (3, 2); open circles at (1, 1) and (3, −1).Open in grapher →

(a) $\displaystyle \lim_{x \to 1^-} F(x)$ (b) $\displaystyle \lim_{x \to 1^+} F(x)$ (c) $\displaystyle \lim_{x \to 1} F(x)$ (d) $\displaystyle \lim_{x \to 3} F(x)$ (e) $F(3)$

Solution.

(a) Coming from the left along the rising line, the heights approach $3$ . So $\displaystyle \lim_{x \to 1^-} F(x) = 3$ .

(b) Coming from the right along the falling line, the heights approach the open circle at height $1$ . So $\displaystyle \lim_{x \to 1^+} F(x) = 1$ .

(d) From both sides of $x = 3$ the falling line approaches the open circle at height $-1$ , so $\displaystyle \lim_{x \to 3} F(x) = -1$ .

(e) The filled dot above $x = 3$ gives $F(3) = 2$ . The value is not the same as the limit.

Recognizing unbounded behavior

When the graph shoots upward or downward next to a vertical asymptote, the outputs do not approach any real number. Consider $y = \dfrac{1}{(x - 1)^2}$ .

y = 1/(x − 1)² rises without bound on both sides of the asymptote x = 1.Open in grapher →

From both sides, the heights increase without bound, so $\displaystyle \lim_{x \to 1} \frac{1}{(x - 1)^2} = \infty$ and the limit does not exist as a real number. If one side went up and the other went down, as with $y = \dfrac{1}{x - 1}$ , you would write the one-sided results separately: $\displaystyle \lim_{x \to 1^-} \frac{1}{x - 1} = -\infty$ and $\displaystyle \lim_{x \to 1^+} \frac{1}{x - 1} = \infty$ .

Worked example: Interpreting a table with different one-sided behavior

The table shows selected values of a function $q$ .

$x$	$1.9$	$1.99$	$1.999$	$2.001$	$2.01$	$2.1$
$q(x)$	$4.81$	$4.98$	$4.998$	$0.003$	$0.03$	$0.3$

What do the values suggest about $\displaystyle \lim_{x \to 2} q(x)$ ?

Solution. From the left, the values are approaching $5$ . From the right, they are approaching $0$ . The table suggests $\displaystyle \lim_{x \to 2^-} q(x) = 5$ and $\displaystyle \lim_{x \to 2^+} q(x) = 0$ , so the two-sided limit does not exist.

Tip

When a graph is drawn on a grid, pause at each open circle and filled dot and write down its coordinates before answering. Most mistakes on graph-reading questions come from mixing up the two.

Practice

Practice 1

The table shows values of $f(x) = \dfrac{\sqrt{x + 4} - 2}{x}$ , rounded to six decimal places.

$x$	$-0.1$	$-0.01$	$-0.001$	$0.001$	$0.01$	$0.1$
$f(x)$	$0.251582$	$0.250156$	$0.250016$	$0.249984$	$0.249844$	$0.248457$

Estimate $\displaystyle \lim_{x \to 0} f(x)$ . Give an exact value.