Estimating Limits from Graphs and Tables

Before you learn the algebra for finding limits exactly, it helps to see them. A graph shows where a function is heading, and a table of values lets you watch the outputs close in on a number. The AP exam gives functions in all these forms (graphs, tables, equations, and words), so you need to read limits from each one.

Key ideas

Limits from a graph

To find $\displaystyle\lim_{x \to a} f(x)$ from a graph:

1. Put your finger on the curve a little to the left of $x = a$ and slide toward $a$. Note the height you approach.
2. Do the same from the right.
3. If both heights are the same number $L$, the limit is $L$. If they differ, or if the curve shoots up or down without bound, the two-sided limit does not exist.

Ignore the dot at $x = a$ while you do this. Filled and open dots tell you $f(a)$, not the limit.

The graph of $g$ used in Example 1 and the practice questions.

Limits from a table

A table can suggest a limit. Choose $x$-values that get closer to $a$ from both sides, such as $a \pm 0.1$, $a \pm 0.01$, $a \pm 0.001$, and look for the number the outputs approach.

A table is only evidence, not proof. It can be fooled if the $x$-values are badly chosen or don’t get close enough (see Example 4 and Practice 9). On the AP exam, tables are often all you get, so say “the values suggest” or “an estimate is”.

Connecting representations

A limit is the same no matter how the function is described, so different forms should agree:

• If a table suggests a limit of $1$, the graph should head toward height $1$.
• If the graph has a hole, the equation usually has a factor that cancels (you’ll do this algebra in algebraic techniques for limits).
• A graphing calculator’s table feature is a quick way to check an answer you found by algebra.

When two representations disagree, look harder: one of them is hiding something.

Worked examples

Example 1: Reading a graph

Use the graph of $g$ above. Find each value, or say why it does not exist.

(a) $\displaystyle\lim_{x \to -2} g(x)$ and $g(-2)$

(b) $\displaystyle\lim_{x \to 0} g(x)$ and $g(0)$

(c) $\displaystyle\lim_{x \to 4} g(x)$

Solution.

(a) From the left, the line rises toward height $3$. From the right, the parabola comes down toward height $1$:

$$\lim_{x \to -2^-} g(x) = 3, \qquad \lim_{x \to -2^+} g(x) = 1$$

They differ, so $\displaystyle\lim_{x \to -2} g(x)$ does not exist. The filled dot gives $g(-2) = 3$.

(b) From both sides the parabola approaches the hole at height $-1$, so $\displaystyle\lim_{x \to 0} g(x) = -1$. The filled dot gives $g(0) = 2$.

(c) Nothing unusual happens at $x = 4$; the curve passes smoothly through the marked point $(4, 0.5)$, so $\displaystyle\lim_{x \to 4} g(x) = 0.5$.

Example 2: Estimating from a table

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

Solution. You can’t substitute $x = 0$ (you’d get $\tfrac{0}{0}$), so use values close to $0$ on both sides (rounded to $6$ decimal places):

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

From the left the values climb toward $1$, and from the right they fall toward $1$. The table suggests

$$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$

(This limit is exactly $1$; it’s the reason $e$ is such a special base.)

Example 3: A table that shows a jump

A function $h$ is given only by this table. Estimate $\displaystyle\lim_{x \to 3^-} h(x)$, $\displaystyle\lim_{x \to 3^+} h(x)$, and $\displaystyle\lim_{x \to 3} h(x)$.

$x$	$2.9$	$2.99$	$2.999$	$3.001$	$3.01$	$3.1$
$h(x)$	$3.81$	$3.98$	$3.998$	$2.501$	$2.51$	$2.6$

Solution. From the left, the values approach $4$. From the right, they approach $2.5$:

$$\lim_{x \to 3^-} h(x) \approx 4, \qquad \lim_{x \to 3^+} h(x) \approx 2.5$$

The estimates disagree, so the table suggests that $\displaystyle\lim_{x \to 3} h(x)$ does not exist.

Example 4: When a table misleads

A student estimates $\displaystyle\lim_{x \to 0} \sin\left(\frac{\pi}{x}\right)$ using $x = 0.1$, $0.01$, and $0.001$ (radians). Every output is $0$, so she concludes the limit is $0$. Is she right?

Solution. Her values give $\sin(10\pi)$, $\sin(100\pi)$, $\sin(1000\pi)$, which are all exactly $0$. But other $x$-values close to $0$ tell a different story. For $x = 0.4$, $\tfrac{2}{9}$, $\tfrac{2}{13}$, $\dots$, the inputs to sine are $\tfrac{5\pi}{2}$, $\tfrac{9\pi}{2}$, $\tfrac{13\pi}{2}$, $\dots$, and the output is $1$ every time.

A graph shows the truth: near $0$ the function oscillates between $-1$ and $1$ infinitely often. The limit does not exist. Her $x$-values all happened to land on zeros of the function. Checking a second representation (the graph) caught the error.

Common mistakes

Reading the dot instead of the curve. In Example 1(b), the dot at $(0, 2)$ is $g(0)$. The limit comes from the curve, which heads to $-1$.

Using a table from one side only. A table with only $x = 2.9, 2.99, 2.999$ tells you about the left-hand limit. You need values from both sides to estimate a two-sided limit.

Trusting a table too much. Tables can mislead (Example 4). If the values jump around or you suspect oscillation, check a graph, and use $x$-values that aren’t “nice” numbers.

Rounding too early. In Example 2, rounding to one decimal place turns the four middle values into $1.0$, so you can no longer see that the left side is below $1$ and the right side is above it. Keep enough decimals to see the trend.

Forgetting radians. Calculus uses radians. A calculator set to degrees gives completely different numbers for trig limits. Check the mode before making a table.

Practice

1. (Warm-up) Use the graph of $g$ above to find $\displaystyle\lim_{x \to -2^+} g(x)$.

Solution

From the right of $x = -2$, the parabola heads toward the open dot at height $1$:

$$\lim_{x \to -2^+} g(x) = 1$$

2. (Warm-up) Use the graph of $g$ to find $\displaystyle\lim_{x \to 0} g(x)$ and $g(0)$. Are they equal?

Solution

$\displaystyle\lim_{x \to 0} g(x) = -1$ (the hole) and $g(0) = 2$ (the filled dot). They are not equal.

3. (Warm-up) Use the table to estimate $\displaystyle\lim_{x \to 3} f(x)$.

$x$	$2.9$	$2.99$	$2.999$	$3.001$	$3.01$	$3.1$
$f(x)$	$6.81$	$6.98$	$6.998$	$7.002$	$7.02$	$7.2$

Solution

From both sides the values approach $7$, so $\displaystyle\lim_{x \to 3} f(x) \approx 7$.

4. (Core) Make a table to estimate $\displaystyle\lim_{x \to 0} \frac{\sin x}{x}$, with $x$ in radians. What would go wrong in degree mode?

Solution

$x$	$\pm 0.1$	$\pm 0.01$	$\pm 0.001$
$\dfrac{\sin x}{x}$	$0.998334$	$0.999983$	$0.9999998$

(The function is even, so both sides give the same values.) The table suggests the limit is $1$.

In degree mode, the calculator would treat $x$ as degrees and the values would approach $\dfrac{\pi}{180} \approx 0.017$ instead. Calculus formulas assume radians.

5. (Core) Make a table to estimate $\displaystyle\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$.

Solution

$x$	$3.9$	$3.99$	$4.01$	$4.1$
$\dfrac{\sqrt{x} - 2}{x - 4}$	$0.251582$	$0.250156$	$0.249844$	$0.248457$

The values approach $0.25$ from both sides, so the limit is about $0.25$, or $\tfrac{1}{4}$.

6. (Core) Use the graph of $g$ to find $\displaystyle\lim_{x \to 2^-} g(x)$ and $\displaystyle\lim_{x \to 2^+} g(x)$. Does $\displaystyle\lim_{x \to 2} g(x)$ exist?

Solution

From the left, the parabola rises to the filled dot at height $1$, so $\displaystyle\lim_{x \to 2^-} g(x) = 1$.

From the right, the curve climbs without bound beside the asymptote, so $\displaystyle\lim_{x \to 2^+} g(x) = \infty$.

The two-sided limit does not exist.

7. (Core) (Calculator) Use a table to estimate $\displaystyle\lim_{x \to 0} (1 + x)^{1/x}$ to three decimal places.

Solution

$x$	$-0.1$	$-0.01$	$-0.001$	$0.001$	$0.01$	$0.1$
$(1 + x)^{1/x}$	$2.867972$	$2.731999$	$2.719642$	$2.716924$	$2.704814$	$2.593742$

The values close in from both sides on about $2.718$. (The exact limit is $e \approx 2.71828$.)

8. (Challenge) Sketch the graph of a function $f$ with all of these features:

• $f(1) = 2$, $\displaystyle\lim_{x \to 1^-} f(x) = 0$, and $\displaystyle\lim_{x \to 1^+} f(x) = 2$
• $\displaystyle\lim_{x \to 3} f(x) = -1$, but $f(3)$ is not defined

Solution

Many graphs work. One example:

• Draw a curve from the left that heads toward $(1, 0)$ and ends with an open dot there.
• Put a filled dot at $(1, 2)$ and start a new curve there going to the right.
• Make that curve pass through an open dot (a hole) at $(3, -1)$ and continue past it, with no dot anywhere else on the line $x = 3$.

Check: at $x = 1$ the left side approaches $0$, the right side approaches $2$, and $f(1) = 2$. At $x = 3$ both sides approach $-1$, but there’s no point on the graph there.

9. (Challenge) Let $f(x) = x^2 - \dfrac{\cos x}{10\,000}$ (radians).

• (a) Make a table of $f(x)$ for $x = 1$, $0.5$, $0.1$, $0.01$. What does it suggest the limit as $x \to 0$ is?
• (b) Find $\displaystyle\lim_{x \to 0} f(x)$ exactly by substituting $x = 0$, which is allowed here since $f$ is made of continuous functions. What went wrong in (a)?

Solution

(a)

$x$	$1$	$0.5$	$0.1$	$0.01$
$f(x)$	$0.999946$	$0.249912$	$0.009900$	$0.000000005$

The values seem to be heading to $0$.

(b) Substituting:

$$\lim_{x \to 0} f(x) = 0 - \frac{\cos 0}{10\,000} = -0.0001$$

The table didn’t get close enough. The tiny term $-\tfrac{\cos x}{10\,000}$ only shows up once $x^2$ is smaller than about $0.0001$. At $x = 0.001$, for example, $f(x) \approx -0.000099$. A table is evidence, not proof.