Mathematics guide, Unit 1, step 9 of 33Previous step Next step

Step 9: Limits from Graphs, Tables and Algebra

By the EconTutorials team. 10-minute read with 6 practice questions.

SyllabusAP 1.1–1.9 · IB SL 5.1

In 30 seconds

$ \lim_{x\to a} f(x) = L $ means $ f(x) $ gets as close to $ L $ as you like when $ x $ is close enough to $ a $, but not equal to it. What happens at $ a $ does not matter.
The limit exists only when the left-hand and right-hand limits exist and are equal. It fails if they differ, if $ f $ grows without bound, or if $ f $ oscillates.
Substitute first. If you get $ \tfrac00 $, factorise and cancel, multiply by a conjugate, or combine fractions, then substitute again. Graphs and tables give estimates; algebra gives proof.
Exam point: the squeeze theorem traps a function between two that share a limit. It proves $ \lim_{x\to0}\frac{\sin x}{x} = 1 $, which is true only with $ x $ in radians.

Video lesson coming soonA short animated lesson for this step will appear here.

Change at an instant

This step opens the calculus units of the guide. It follows AP Calculus Unit 1, Limits and Continuity, topics 1.1 to 1.9; the next step covers topics 1.10 to 1.16. In IB Analysis and Approaches the limit is introduced at the start of topic 5 (SL 5.1). Cambridge 9709 has no separate limits topic, but Paper 1 asks you to see a gradient as the limit of chord gradients, which is exactly the algebra on this page.

Why do we need limits at all? Speed is distance divided by time. At a single instant both are zero, and $ \tfrac00 $ means nothing. Yet a speedometer shows a number. A stone dropped from a cliff falls $ s = 4t^2 $ metres in $ t $ seconds. Between $ t = 3 $ and $ t = 3 + h $ its average speed is

\[ \frac{4(3 + h)^2-36}{h} = \frac{24h + 4h^2}{h} = 24 + 4h \quad (h \neq 0). \]

For $ h = 0.1 $, 0.01 and 0.001 this is 24.4, 24.04 and 24.004 m/s. We cannot put $ h = 0 $ in the original fraction, but as $ h $ shrinks the average speed closes in on 24. That number, 24 m/s, is the speed at the instant $ t = 3 $. It is a limit, and every derivative in the next unit is built the same way.

Key termLimit: $ \lim_{x\to a} f(x) = L $ means that $ f(x) $ can be made as close to $ L $ as we like by taking $ x $ close enough to $ a $, with $ x \neq a $.

Take $ f(x) = \dfrac{x^2-9}{x-3} $. At $ x = 3 $ it gives $ \tfrac00 $, so $ f(3) $ does not exist. Everywhere else the top factorises and $ f(x) = x + 3 $. So the graph is the line $ y = x + 3 $ with one point missing, and as $ x $ approaches 3 from either side, $ f(x) $ approaches 6. Hence $ \lim_{x\to3} f(x) = 6 $, even though $ f(3) $ is undefined. A limit describes the approach, never the arrival.

Sometimes the two sides disagree. We write $ \lim_{x\to a^-} f(x) $ for the limit from the left ($ x < a $) and $ \lim_{x\to a^+} f(x) $ for the limit from the right ($ x > a $).

\[ \lim_{x\to a} f(x) = L \iff \lim_{x\to a^-} f(x) = L = \lim_{x\to a^+} f(x) \]

Reading limits from graphs and tables

To read a limit from a graph, trace the curve towards $ x = a $ from each side with a finger and see what height you approach. Open circles and filled dots tell you about $ f(a) $, which the limit ignores.

Reading limits from a graph: y = f(x)

Figure 1. f(x) = x + 1 for x < 1 and f(x) = 4 − x for x > 1, with f(1) = 1 and f(3) = 3. Open circles are points missing from the graph; filled dots are the actual values.

Worked example

For the function in Figure 1, find (a) $ \lim_{x\to1^-} f(x) $ (b) $ \lim_{x\to1^+} f(x) $ (c) $ \lim_{x\to1} f(x) $ (d) $ f(1) $ (e) $ \lim_{x\to3} f(x) $ and $ f(3) $.

Step 1(a) From the left along $ y = x + 1 $, the height approaches 2.

Step 2(b) From the right along $ y = 4-x $, the height approaches 3.

Step 3(c) The one-sided limits differ, so $ \lim_{x\to1} f(x) $ does not exist.

Step 4(d) The filled dot gives $ f(1) = 1 $, a value that matches neither side.

Step 5(e) Near $ x = 3 $ both sides follow $ y = 4-x $ towards height 1, so the limit is 1. But the filled dot says $ f(3) = 3 $.

Part (e) is the key idea: the limit and the value can be different numbers.

A limit can fail to exist in three ways. The one-sided limits differ, as at $ x = 1 $ in Figure 1. The function grows without bound: $ \frac{1}{(x-2)^2} $ shoots up as $ x \to 2 $. We may write $ \lim_{x\to2} \frac{1}{(x-2)^2} = \infty $ to say how it fails, but $ \infty $ is not a number, so the limit still does not exist. Or the function oscillates: $ \sin\frac1x $ swings between $ -1 $ and 1 infinitely often as $ x \to 0 $ and settles on nothing.

When there is no graph, a table of values close to $ a $ on both sides gives an estimate.

Worked example

Estimate $ \displaystyle\lim_{x\to0} \frac{e^x-1}{x} $ from a table of values.

Step 1Substitution gives $ \tfrac{0}{0} $, so evaluate either side of 0 instead.

Step 2Values to 4 decimal places are shown in the table below.

Step 3From both sides the values approach 1, so the limit appears to be 1.

$ x $	−0.1	−0.01	−0.001	0.001	0.01	0.1
$ \frac{e^x-1}{x} $	0.9516	0.9950	0.9995	1.0005	1.0050	1.0517

No algebra trick removes the $ \tfrac00 $ here, so the table is genuinely useful. L’Hôpital’s rule, in Step 16, will prove the value is exactly 1.

Common mistakeTrusting a table blindly. For $ f(x) = \cos\frac{\pi}{x} $, the inputs $ x = 0.1, 0.01, 0.001 $ give $ \cos 10\pi = \cos 100\pi = \cos 1000\pi = 1 $, which suggests a limit of 1. Yet $ x = 0.2 $ gives $ \cos 5\pi = -1 $. The function swings between $ -1 $ and 1 all the way to 0, and the limit does not exist. A table suggests; algebra confirms.

Evaluating limits with algebra

Limits behave well with arithmetic. If $ \lim_{x\to a} f(x) = L $ and $ \lim_{x\to a} g(x) = M $, then the limit of a sum is $ L + M $, of a product is $ LM $, of $ kf(x) $ is $ kL $, of a power is $ L^n $, and of a quotient is $ \frac{L}{M} $ provided $ M \neq 0 $. These are the limit laws.

They give the most useful rule of all, direct substitution: for a polynomial, $ \lim_{x\to a} p(x) = p(a) $, and the same holds for a rational function wherever the denominator is not zero. So always substitute first.

Worked example

Given $ \lim_{x\to4} f(x) = 3 $ and $ \lim_{x\to4} g(x) = -1 $, find $ \displaystyle\lim_{x\to4} \frac{[f(x)]^2-2g(x)}{f(x) + g(x)} $.

Step 1Top: $ 3^2-2(-1) = 11 $.

Step 2Bottom: $ 3 + (-1) = 2 $, which is not zero, so the quotient law applies.

Step 3The limit is $ \tfrac{11}{2} $.

Check the denominator’s limit before you divide. If it is zero, the quotient law says nothing.

When substitution gives $ \tfrac00 $, the function is hiding a common factor. Why may we cancel it? Because $ \frac{(x-2)(x^2 + 2x + 4)}{(x-2)(x + 2)} $ and $ \frac{x^2 + 2x + 4}{x + 2} $ are equal at every $ x $ except 2, and a limit as $ x \to 2 $ never looks at $ x = 2 $ itself. So the two have the same limit. Here are the three standard repairs.

Worked example

Find (a) $ \displaystyle\lim_{x\to2} \frac{x^3-8}{x^2-4} $ (b) $ \displaystyle\lim_{x\to5} \frac{\sqrt{x + 11}-4}{x-5} $ (c) $ \displaystyle\lim_{x\to0} \frac{1}{x}\left(\frac{1}{x + 2}-\frac12\right) $.

Step 1(a) Factorise: $ \dfrac{(x-2)(x^2 + 2x + 4)}{(x-2)(x + 2)} = \dfrac{x^2 + 2x + 4}{x + 2} \to \dfrac{12}{4} = 3 $.

Step 2(b) Multiply top and bottom by the conjugate $ \sqrt{x + 11} + 4 $. The top becomes $ (x + 11)-16 = x-5 $.

Step 3So the fraction is $ \dfrac{1}{\sqrt{x + 11} + 4} \to \dfrac{1}{4 + 4} = \dfrac18 $.

Step 4(c) Combine: $ \dfrac{1}{x + 2}-\dfrac12 = \dfrac{2-(x + 2)}{2(x + 2)} = \dfrac{-x}{2(x + 2)} $.

Step 5Divide by $ x $: $ \dfrac{-1}{2(x + 2)} \to -\dfrac14 $.

Factorise for polynomials, use the conjugate for square roots, and combine compound fractions over a common denominator.

Substitution gives	What it means	What to do
a real number	that is the limit (for standard functions)	stop
$ \frac{k}{0} $, $ k \neq 0 $	unbounded near $ a $	check the sign on each side
$ \frac00 $	indeterminate: anything is possible	factorise, conjugate, combine fractions, or use $ \frac{\sin x}{x} \to 1 $
piecewise or $ \|x\| $	the two sides may differ	find each one-sided limit

An economics example: marginal cost is a limit. A firm’s total cost is $ TC(Q) = 50 + 6Q + 0.2Q^2 $ dollars. Raising output from 25 to $ 25 + h $ units costs, per extra unit,

\[ \frac{TC(25 + h)-TC(25)}{h} = 16 + 0.2h \]

because the top simplifies to $ 16h + 0.2h^2 $.

For $ h = 1 $ (the 26th unit) that is \$16.20. When output is large, one unit is a tiny step, so economists take the limit as $ h \to 0 $: the marginal cost at $ Q = 25 $ is \$16. Whenever an economist says “at the margin”, a limit is hiding underneath.

The squeeze theorem and $ \sin x / x $

Some limits resist algebra. Consider $ x^2\cos\frac1x $ as $ x \to 0 $. The factor $ \cos\frac1x $ oscillates and has no limit. But it is trapped between $ -1 $ and 1, and it is multiplied by $ x^2 $, which shrinks to 0. Trap a function between two simpler ones that meet, and it has nowhere else to go.

Key termSqueeze theorem: if $ g(x) \le f(x) \le h(x) $ for all $ x $ near $ a $ (except possibly at $ a $), and $ \lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L $, then $ \lim_{x\to a} f(x) = L $.

Worked example

Show that $ \displaystyle\lim_{x\to0} x^2\cos\tfrac1x = 0 $.

Step 1For every $ x \neq 0 $, $ -1 \le \cos\frac1x \le 1 $.

Step 2Multiply by $ x^2 $, which is positive, so the inequalities keep their direction: $ -x^2 \le x^2\cos\frac1x \le x^2 $.

Step 3Both $ -x^2 $ and $ x^2 $ tend to 0 as $ x \to 0 $.

Step 4By the squeeze theorem, $ \lim_{x\to0} x^2\cos\frac1x = 0 $.

Name the theorem and show both bounding limits; that is what AP readers look for.

Squeezing y = x² cos(1/x)

y = x²y = −x²y = x² cos(1/x)

Figure 2. x² cos(1/x) wiggles faster and faster near 0, but it is trapped between y = −x² and y = x², which both pass through the origin. So it is squeezed to 0.

The most famous squeeze is this one, with $ x $ in radians.

\[ \lim_{x\to0} \frac{\sin x}{x} = 1 \qquad\qquad \lim_{x\to0} \frac{1-\cos x}{x} = 0 \]

In a unit circle, compare three nested shapes for an angle $ 0 < x < \frac{\pi}{2} $: a triangle of area $ \frac12\sin x $, a sector of area $ \frac12 x $, and a larger triangle of area $ \frac12\tan x $. Their areas are in order, which rearranges to $ \cos x < \frac{\sin x}{x} < 1 $. As $ x \to 0 $, $ \cos x \to 1 $, so $ \frac{\sin x}{x} $ is squeezed to 1 (Figure 3). The sector formula $ \frac12 r^2\theta $ is a radian formula, so the result holds only in radians; in degrees the limit is $ \frac{\pi}{180} $.

y = (sin x)/x, x in radians

y = (sin x)/x

Figure 3. (sin x)/x is undefined at x = 0, but from both sides the curve heads to height 1. The open circle marks the limit, which is 1.

Worked example

Find (a) $ \displaystyle\lim_{x\to0} \frac{\sin 3x}{5x} $ (b) $ \displaystyle\lim_{x\to0} \frac{1-\cos 2x}{x^2} $.

Step 1(a) The pattern needs the same quantity on top and bottom. Write $ \dfrac{\sin 3x}{5x} = \dfrac35 \cdot \dfrac{\sin 3x}{3x} $.

Step 2As $ x \to 0 $, $ 3x \to 0 $, so $ \dfrac{\sin 3x}{3x} \to 1 $ and the limit is $ \dfrac35 $.

Step 3(b) From Step 8, $ 1-\cos 2x = 2\sin^2 x $, so the fraction is $ 2\left(\dfrac{\sin x}{x}\right)^2 $.

Step 4The limit is $ 2 \times 1^2 = 2 $.

A trigonometric identity often turns an awkward limit into the standard one.

Exam tipAP often mixes representations in one question: $ f $ given by a graph, $ g $ by a table, $ h $ by a formula. Find each limit in its own representation, then combine them with the limit laws. Keep writing $ \lim_{x\to a} $ on every line until the moment you substitute, and when you use the squeeze theorem, state the inequality, both bounding limits and the theorem’s name.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. Find $ \displaystyle\lim_{x\to5} \frac{x^2-25}{x-5} $. Is the function defined at $ x = 5 $? Show worked answer

Substitution gives $ \frac00 $, so factorise: $ \frac{(x-5)(x + 5)}{x-5} = x + 5 $ for $ x \neq 5 $.
The limit is $ 5 + 5 = 10 $.
The function is not defined at $ x = 5 $: the graph has a hole at $ (5, 10) $.

EasyQ2. $ f(x) = 2x-1 $ for $ x < 3 $ and $ f(x) = x^2-3 $ for $ x \ge 3 $. Find both one-sided limits at $ x = 3 $. Does $ \lim_{x\to3} f(x) $ exist? Show worked answer

Left: $ \lim_{x\to3^-} (2x-1) = 5 $.
Right: $ \lim_{x\to3^+} (x^2-3) = 6 $.
They differ, so $ \lim_{x\to3} f(x) $ does not exist.

MediumQ3. Find $ \displaystyle\lim_{x\to-2} \frac{x^2 + 5x + 6}{x^2-4} $. Show worked answer

Substitution gives $ \frac{4-10 + 6}{4-4} = \frac00 $.
Factorise: $ \frac{(x + 2)(x + 3)}{(x + 2)(x-2)} = \frac{x + 3}{x-2} $ for $ x \neq -2 $.
Substitute: $ \frac{1}{-4} = -\frac14 $.

MediumQ4. Find $ \displaystyle\lim_{x\to0} \frac{\sqrt{9 + x}-3}{x} $. Show worked answer

Multiply top and bottom by $ \sqrt{9 + x} + 3 $.
Top: $ (9 + x)-9 = x $, so the fraction is $ \frac{1}{\sqrt{9 + x} + 3} $ for $ x \neq 0 $.
Substitute: $ \frac{1}{3 + 3} = \frac16 $.

MediumQ5. Find $ \displaystyle\lim_{x\to0} \frac{\tan 2x}{3x} $. Show worked answer

$ \frac{\tan 2x}{3x} = \frac{\sin 2x}{3x\cos 2x} = \frac23 \cdot \frac{\sin 2x}{2x} \cdot \frac{1}{\cos 2x} $.
As $ x \to 0 $: $ \frac{\sin 2x}{2x} \to 1 $ and $ \cos 2x \to 1 $.
The limit is $ \frac23 $.

ExamQ6. (AP style) The function $ h $ satisfies $ 3x-2 \le h(x) \le x^2 + x-1 $ for all $ x $. Let $ k(x) = \dfrac{x^2-1}{x-1} $ for $ x \neq 1 $. (a) Find $ \lim_{x\to1} h(x) $. Justify your answer. [2] (b) Find $ \lim_{x\to1} h(x)\,k(x) $. [2] (c) Find $ \displaystyle\lim_{x\to1} \frac{\sin(\pi x-\pi)}{x-1} $. [2] Show worked answer

(a) $ \lim_{x\to1}(3x-2) = 1 $ and $ \lim_{x\to1}(x^2 + x-1) = 1 $ [1]. Since $ h $ is squeezed between them, $ \lim_{x\to1} h(x) = 1 $ by the squeeze theorem [1].
(b) $ k(x) = x + 1 $ for $ x \neq 1 $, so $ \lim_{x\to1} k(x) = 2 $ [1]. By the product law the limit is $ 1 \times 2 = 2 $ [1].
(c) Let $ u = \pi(x-1) $; as $ x \to 1 $, $ u \to 0 $. The fraction is $ \pi \cdot \frac{\sin u}{u} $ [1], so the limit is $ \pi $ [1].
Total: 6 marks.

Check yourself

1. $ \lim_{x\to2^-} f(x) = 4 $, $ \lim_{x\to2^+} f(x) = 4 $ and $ f(2) = 7 $. What is $ \lim_{x\to2} f(x) $?

Both one-sided limits equal 4, so the limit is 4. The value $ f(2) = 7 $ does not affect the limit at all.

2. Direct substitution into a limit gives $ \frac{5}{0} $. What can you conclude?

A non-zero number divided by something tending to 0 grows without bound. Check the sign on each side to see whether it heads to $ +\infty $ or $ -\infty $.

3. What is $ \lim_{x\to0} \frac{\sin 4x}{x} $, with $ x $ in radians?

$ \frac{\sin 4x}{x} = 4 \cdot \frac{\sin 4x}{4x} \to 4 \times 1 = 4 $.

Your score: 0 out of 3

The short version

A limit describes what $ f(x) $ approaches as $ x \to a $, never what happens at $ a $. It exists only when both one-sided limits exist and agree.
Limits fail to exist when the sides differ, when the function is unbounded, or when it oscillates.
Graphs and tables estimate limits; a table can mislead if the inputs are badly chosen.
Substitute first. For $ \frac00 $, factorise, use the conjugate or combine fractions, then substitute again.
The squeeze theorem traps a function between two with the same limit; it gives $ \frac{\sin x}{x} \to 1 $ in radians.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

Step 9: Limits from Graphs, Tables and Algebra

Change at an instant

Reading limits from graphs and tables

Evaluating limits with algebra

The squeeze theorem and \( \sin x / x \)

Practise: 6 questions

Check yourself

The short version

Further reading on EconTutorials

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