Mathematics guide, Unit 4, step 16 of 33Previous step Next step

Step 16: Linearisation and L'Hôpital's Rule

By the EconTutorials team. 9-minute read with 6 practice questions.

SyllabusAP 4.6–4.7 · IB HL 5.13

In 30 seconds

Zoom in far enough on a differentiable curve and it looks straight. Near $ x = a $, the curve is close to its tangent, the linearisation $ L(x) = f(a) + f'(a)(x-a) $, so $ f(x) \approx L(x) $.
Concavity decides the direction of the error. If the curve is concave down near $ a $, it lies below its tangent and $ L(x) $ is an overestimate; concave up gives an underestimate.
L’Hôpital’s rule: if $ \dfrac{f(x)}{g(x)} $ has the form $ \dfrac{0}{0} $ or $ \dfrac{\infty}{\infty} $, then $ \displaystyle\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)} $, provided the second limit exists. Differentiate the top and the bottom separately.
Exam point: always show that the limit is indeterminate before using the rule. If substitution gives a number, that number is the limit and L’Hôpital’s rule does not apply.

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

Local linearity

Plot $ y = \sqrt[3]{x} $ and zoom in on the point $ (27, 3) $. Zoom again, and again. The curve straightens out until it is impossible to tell apart from its tangent line. Every differentiable function behaves like this, and it is the reason the derivative exists at all: close to a point, the function is very nearly linear, and the derivative is the gradient of that line.

That gives a practical tool. Near $ x = a $, replace the curve by its tangent. The tangent passes through $ \big(a, f(a)\big) $ with gradient $ f'(a) $ (Step 13), so

\[ f(x) \approx L(x) = f(a) + f'(a)(x-a) \qquad \text{for } x \text{ close to } a. \]

Key termLinearisation (also tangent-line or linear approximation): the function $ L(x) = f(a) + f'(a)(x-a) $, whose graph is the tangent to $ y = f(x) $ at $ x = a $. It approximates $ f(x) $ well when $ x $ is close to $ a $.

In words: new value $ \approx $ old value $ + $ rate $ \times $ change. You met the same idea in Step 15, when the cake’s temperature half a minute later was estimated from its rate of cooling.

Worked example

Use the linearisation of $ f(x) = \sqrt[3]{x} $ at $ x = 27 $ to estimate $ \sqrt[3]{27.6} $. Is the estimate too big or too small?

Step 1Choose $ a = 27 $, close to 27.6 and with an easy cube root: $ f(27) = 3 $.

Step 2$ f(x) = x^{1/3} $, so $ f'(x) = \tfrac{1}{3}x^{-2/3} $ and $ f'(27) = \tfrac{1}{3} \times \tfrac{1}{9} = \tfrac{1}{27} $.

Step 3$ L(x) = 3 + \tfrac{1}{27}(x-27) $, so $ \sqrt[3]{27.6} \approx L(27.6) = 3 + \tfrac{0.6}{27} \approx 3.0222 $.

Step 4$ f”(x) = -\tfrac{2}{9}x^{-5/3} < 0 $ for $ x > 0 $, so the curve is concave down and lies below its tangent: the estimate is too big.

A calculator gives $ \sqrt[3]{27.6} \approx 3.02206 $, so the estimate is high by only about 0.0002 (Figure 1).

y = ∛x and its tangent at x = 27

y = ∛xtangent L(x) = 3 + (x − 27)/27

Figure 1. Near x = 27 the tangent and the curve are almost indistinguishable, so L(27.6) = 3.0222 is a good estimate of ∛27.6. The curve is concave down, so it bends away below its tangent and the estimate is slightly too big.

Over or under? Reading the concavity

The tangent line ignores curvature, and the second derivative measures curvature. That is all you need to decide the direction of the error.

Near $ x = a $	Shape	Tangent lies	$ L(x) $ is
$ f”(x) > 0 $	Concave up (bends upwards)	Below the curve	An underestimate
$ f”(x) < 0 $	Concave down (bends downwards)	Above the curve	An overestimate

How good is the estimate? Roughly, the error grows with the square of the distance from $ a $: step twice as far away and the error is about four times as big. For $ \sqrt[3]{x} $, moving from 27 to 27.6 gave an error of about 0.0002, while moving to 33 gives $ L(33) \approx 3.2222 $ against a true value of 3.2075, an error of about 0.015. Stay close to $ a $.

AP free-response questions often give no formula at all, only values of $ f $ and $ f’ $ and some information about $ f” $.

Worked example

A differentiable function $ f $ has $ f(2) = 5 $ and $ f'(2) = -3 $, and $ f”(x) > 0 $ for all $ x $. Use the tangent line at $ x = 2 $ to estimate $ f(2.1) $, and say whether the estimate is too big or too small.

Step 1Tangent line: $ L(x) = 5-3(x-2) $.

Step 2$ f(2.1) \approx L(2.1) = 5-3(0.1) = 4.7 $.

Step 3$ f” > 0 $, so the graph of $ f $ is concave up and lies above its tangent line. The estimate 4.7 is an underestimate.

The answer to “too big or too small?” must give the reason: the sign of $ f” $, the concavity, and where the tangent lies.

An economics example: the marginal product. A firm’s output is $ Q = 20\sqrt{L} $ units when it employs $ L $ workers. At $ L = 100 $, $ Q = 200 $, and the marginal product of labour is $ \dfrac{dQ}{dL} = \dfrac{10}{\sqrt{L}} = 1 $ unit per worker. Linearisation predicts that hiring 3 more workers adds about $ 1 \times 3 = 3 $ units. The exact gain is $ 20\sqrt{103}-200 \approx 2.978 $ units. The estimate is a little high because $ Q $ is concave down: each extra worker adds slightly less than the one before, the law of diminishing returns. Marginal cost, marginal revenue and marginal product are all linearisations in disguise.

Small changes. The same idea, written with the change $ \delta x = x-a $, says that a small change in $ x $ produces a change in $ y $ of about $ \delta y \approx \dfrac{dy}{dx}\,\delta x $. This is how scientists and engineers estimate the effect of a measurement error.

Worked example

The radius of a ball is measured as 10 cm, with a possible error of 0.1 cm. Estimate the possible error in its calculated volume, and express it as a percentage.

Step 1$ V = \tfrac{4}{3}\pi r^3 $, so $ \dfrac{dV}{dr} = 4\pi r^2 $, which is $ 400\pi $ at $ r = 10 $.

Step 2$ \delta V \approx 400\pi \times 0.1 = 40\pi \approx 126 $ cm³.

Step 3$ V = \tfrac{4000\pi}{3} \approx 4189 $ cm³, so the percentage error is about $ \tfrac{40\pi}{4000\pi/3} \times 100 = 3 $ per cent.

A 1% error in the radius becomes about a 3% error in the volume, because the volume depends on $ r^3 $. Each power multiplies the percentage error.

Common mistakeChoosing a base point that is not close to $ x $, or one where $ f(a) $ is not easy to find. To estimate $ \sqrt{50} $, use $ a = 49 $, not $ a = 36 $ or $ a = 50 $. And keep the brackets: $ L(x) = f(a) + f'(a)(x-a) $, so the change $ x-a $ can be negative, as in $ L(2.1) $ above, where the gradient was negative.

Indeterminate forms

In Step 9 you settled limits such as $ \displaystyle\lim_{x \to 2}\frac{x^2-4}{x-2} $ by factorising. Substituting first gives $ \dfrac{0}{0} $, which is not a number but a signal that more work is needed. Some limits of this kind will not factorise at all: $ \dfrac{1-\cos x}{x^2} $ as $ x \to 0 $, or $ \dfrac{x^3}{e^{2x}} $ as $ x \to \infty $.

Key termIndeterminate form: a limit whose direct substitution gives $ \dfrac{0}{0} $ or $ \dfrac{\infty}{\infty} $ (or a related form such as $ 0 \cdot \infty $ or $ 1^{\infty} $). The form alone does not tell you the value; the limit could be any number, or infinite.

Key termL’Hôpital’s rule: suppose $ f $ and $ g $ are differentiable near $ a $ (except possibly at $ a $), $ g'(x) \neq 0 $ there, and $ \displaystyle\lim_{x \to a}\frac{f(x)}{g(x)} $ has the form $ \dfrac{0}{0} $ or $ \dfrac{\pm\infty}{\pm\infty} $. Then $ \displaystyle\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} $, provided the right-hand limit exists or is $ \pm\infty $. The same holds for one-sided limits and as $ x \to \pm\infty $.

Why should a ratio of functions match a ratio of derivatives? Local linearity again. If $ f(a) = g(a) = 0 $, then near $ a $, $ f(x) \approx f'(a)(x-a) $ and $ g(x) \approx g'(a)(x-a) $. The small factor $ x-a $ cancels, leaving $ \dfrac{f(x)}{g(x)} \approx \dfrac{f'(a)}{g'(a)} $. When two functions race to zero, the one that gets there faster wins, and the derivative measures how fast.

Using L’Hôpital’s rule

The routine is short: check the form, differentiate the top and the bottom separately, and try the limit again. If the new limit is still indeterminate, apply the rule again.

Worked example

Find (a) $ \displaystyle\lim_{x \to 0}\frac{1-\cos x}{x^2} $ (b) $ \displaystyle\lim_{x \to \infty}\frac{x^3}{e^{2x}} $ (c) $ \displaystyle\lim_{x \to 2}\frac{x^3-8}{x^2-4} $.

Step 1(a) Form $ \dfrac{0}{0} $. Differentiate top and bottom: $ \dfrac{\sin x}{2x} $, still $ \dfrac{0}{0} $. Again: $ \dfrac{\cos x}{2} \to \dfrac{1}{2} $. The limit is $ \tfrac{1}{2} $ (Figure 2).

Step 2(b) Form $ \dfrac{\infty}{\infty} $. Three applications: $ \dfrac{3x^2}{2e^{2x}} $, then $ \dfrac{6x}{4e^{2x}} $, then $ \dfrac{6}{8e^{2x}} \to 0 $. The limit is 0: any exponential growth beats any power of $ x $.

Step 3(c) Form $ \dfrac{0}{0} $. $ \dfrac{3x^2}{2x} \to \dfrac{12}{4} = 3 $.

Step 4Check (c) by factorising: $ \dfrac{(x-2)(x^2 + 2x + 4)}{(x-2)(x + 2)} = \dfrac{x^2 + 2x + 4}{x + 2} \to \dfrac{12}{4} = 3 $. The two methods agree.

Write down the form, $ \dfrac{0}{0} $ or $ \dfrac{\infty}{\infty} $, every time you apply the rule, including the second and third time.

y = (1 − cos x)/x² near x = 0

y = (1 − cos x)/x²

Figure 2. Substituting x = 0 gives 0/0, but the graph heads clearly for 1/2 from both sides. L’Hôpital’s rule, used twice, confirms the value. The hollow circle marks the missing point.

Common mistakeTwo classic slips. First, using the quotient rule: L’Hôpital’s rule differentiates the top and the bottom separately, giving $ \dfrac{f’}{g’} $, not $ \left(\dfrac{f}{g}\right)’ $. Second, applying the rule when the form is not indeterminate. $ \displaystyle\lim_{x \to 0}\frac{x + 1}{x + 3} = \frac{1}{3} $ by substitution, but differentiating would wrongly give 1.

Other forms. A product $ 0 \cdot \infty $ must first be rewritten as a quotient. For $ \displaystyle\lim_{x \to 0^+}\sqrt{x}\ln x $, write $ \dfrac{\ln x}{x^{-1/2}} $, which has the form $ \dfrac{-\infty}{\infty} $. Then $ \dfrac{1/x}{-\tfrac{1}{2}x^{-3/2}} = -2\sqrt{x} \to 0 $. A power such as $ 1^{\infty} $ is tamed by taking logarithms first, as the next example shows. AP requires only the $ \dfrac{0}{0} $ and $ \dfrac{\infty}{\infty} $ forms; the rewrites are a useful extra.

Worked example

An economics example: continuous compounding. $1,000 is invested at an annual rate of 5%, compounded $ n $ times a year, so after one year it is worth $ 1000\left(1 + \dfrac{0.05}{n}\right)^n $. What happens as $ n \to \infty $?

Step 1The base tends to 1 and the power to $ \infty $: the form $ 1^{\infty} $. Let $ y = \left(1 + \dfrac{0.05}{n}\right)^n $, so $ \ln y = n\ln\left(1 + \dfrac{0.05}{n}\right) = \dfrac{\ln(1 + 0.05/n)}{1/n} $, the form $ \dfrac{0}{0} $.

Step 2Differentiate top and bottom with respect to $ n $: $ \dfrac{\dfrac{-0.05/n^2}{1 + 0.05/n}}{-1/n^2} = \dfrac{0.05}{1 + 0.05/n} \to 0.05 $.

Step 3So $ \ln y \to 0.05 $ and $ y \to e^{0.05} $. The investment tends to $ 1000e^{0.05} \approx \$1051.27 $.

Annual compounding gives $1,050.00 and monthly compounding $1,051.16, so compounding ever more often adds less and less, and the limit is continuous compounding, $ Pe^{rt} $.

Exam tipLinearisation is AP topic 4.6 and L’Hôpital’s rule is AP topic 4.7; both are on the AB and the BC exams. IB HL 5.13 includes L’Hôpital’s rule, with repeated use. Cambridge 9709 does not include L’Hôpital’s rule. AP questions often give the limit through a table: if $ f(3) = g(3) = 0 $, the limit of $ \dfrac{f(x)}{g(x)} $ as $ x \to 3 $ is $ \dfrac{f'(3)}{g'(3)} $, provided $ g'(3) \neq 0 $ and $ f’ $, $ g’ $ are continuous there.

Practise: 6 questions

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

EasyQ1. Find the linearisation of $ f(x) = \dfrac{1}{x} $ at $ x = 2 $, and use it to estimate $ \dfrac{1}{2.04} $. Show worked answer

$ f(2) = \tfrac{1}{2} $, $ f'(x) = -\dfrac{1}{x^2} $, so $ f'(2) = -\tfrac{1}{4} $.
$ L(x) = \tfrac{1}{2}-\tfrac{1}{4}(x-2) $.
$ L(2.04) = 0.5-0.25 \times 0.04 = 0.49 $.
(True value 0.4902. $ f” = \dfrac{2}{x^3} > 0 $, so this is an underestimate.)

EasyQ2. Find $ \displaystyle\lim_{x \to 0}\frac{\sin 5x}{2x} $. Show worked answer

Form $ \dfrac{0}{0} $.
L’Hôpital: $ \dfrac{5\cos 5x}{2} \to \dfrac{5}{2} $.

MediumQ3. Find $ \displaystyle\lim_{x \to 0}\frac{e^{3x}-1-3x}{x^2} $. Show worked answer

Form $ \dfrac{0}{0} $. Differentiate: $ \dfrac{3e^{3x}-3}{2x} $, still $ \dfrac{0}{0} $.
Again: $ \dfrac{9e^{3x}}{2} \to \dfrac{9}{2} $.

MediumQ4. A function $ f $ has $ f(1) = 3 $, $ f'(1) = -2 $ and $ f”(x) < 0 $ for all $ x $. Use the tangent line at $ x = 1 $ to estimate $ f(0.8) $. Is your estimate too big or too small? Show worked answer

$ L(x) = 3-2(x-1) $.
$ f(0.8) \approx 3-2(-0.2) = 3.4 $.
$ f” < 0 $: the graph is concave down and lies below its tangent, so 3.4 is too big (an overestimate).

MediumQ5. Find $ \displaystyle\lim_{x \to \infty}\frac{(\ln x)^2}{x} $. Show worked answer

Form $ \dfrac{\infty}{\infty} $. Differentiate: $ \dfrac{2\ln x \cdot \frac{1}{x}}{1} = \dfrac{2\ln x}{x} $, still $ \dfrac{\infty}{\infty} $.
Again: $ \dfrac{2/x}{1} = \dfrac{2}{x} \to 0 $.
The limit is 0: even the square of a logarithm grows more slowly than $ x $.

ExamQ6. Let $ f(x) = \ln(1 + 2x) $ for $ x > -\tfrac{1}{2} $. (a) Find the equation of the tangent line to the graph of $ f $ at $ x = 0 $, and use it to approximate $ f(0.1) $. [2] (b) Is the approximation in (a) an overestimate or an underestimate? Give a reason. [2] (c) Use L’Hôpital’s rule to find $ \displaystyle\lim_{x \to 0}\frac{f(x)-2x}{x^2} $. [3] Show worked answer

(a) $ f(0) = 0 $, $ f'(x) = \dfrac{2}{1 + 2x} $, so $ f'(0) = 2 $ and the tangent is $ y = 2x $. [1]
$ f(0.1) \approx 2(0.1) = 0.2 $. [1]
(b) $ f”(x) = -\dfrac{4}{(1 + 2x)^2} < 0 $, so the graph is concave down and lies below the tangent. [1]
The approximation is an overestimate. (True value $ \ln 1.2 \approx 0.182 $.) [1]
(c) Form $ \dfrac{0}{0} $, since $ f(0)-0 = 0 $. Differentiate: $ \dfrac{\frac{2}{1 + 2x}-2}{2x} $, still $ \dfrac{0}{0} $. [1]
Again: $ \dfrac{-\frac{4}{(1 + 2x)^2}}{2} $. [1]
As $ x \to 0 $ this tends to $ -\dfrac{4}{2} = -2 $. [1]
Total: 7 points.

Check yourself

1. $ f(3) = g(3) = 0 $, $ f'(3) = 4 $ and $ g'(3) = -2 $, with $ f’ $ and $ g’ $ continuous. What is $ \displaystyle\lim_{x \to 3}\frac{f(x)}{g(x)} $?

The form is $ \dfrac{0}{0} $, so the limit is $ \dfrac{f'(3)}{g'(3)} = \dfrac{4}{-2} = -2 $.

2. What is $ \displaystyle\lim_{x \to 0}\frac{x + 1}{x + 3} $?

Substitution gives $ \dfrac{1}{3} $, a number, so that is the limit. The form is not indeterminate, so L’Hôpital’s rule does not apply; using it would wrongly give 1.

3. $ f(4) = 10 $ and $ f'(4) = 3 $. What does the linearisation at $ x = 4 $ give for $ f(4.2) $?

$ L(4.2) = 10 + 3(0.2) = 10.6 $.

Your score: 0 out of 3

The short version

Linearisation: $ f(x) \approx L(x) = f(a) + f'(a)(x-a) $ for $ x $ near $ a $. Choose $ a $ close to $ x $ with $ f(a) $ easy to find.
$ f” > 0 $ (concave up): the tangent lies below, so $ L $ underestimates. $ f” < 0 $ (concave down): $ L $ overestimates.
L’Hôpital’s rule: for the forms $ \dfrac{0}{0} $ and $ \dfrac{\infty}{\infty} $, $ \lim\dfrac{f}{g} = \lim\dfrac{f’}{g’} $. Differentiate top and bottom separately, and repeat if needed.
Check the form first. Rewrite $ 0 \cdot \infty $ as a quotient, and take logarithms for $ 1^{\infty} $, as in $ \left(1 + \dfrac{r}{n}\right)^n \to e^{r} $.
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 16: Linearisation and L'Hôpital's Rule

Local linearity

Over or under? Reading the concavity

Indeterminate forms

Using L’Hôpital’s rule

Practise: 6 questions

Check yourself

The short version

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