5.13 Limits and l'Hôpital's rule (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

Indeterminate forms

Substituting $x = a$ into $\dfrac{f(x)}{g(x)}$ sometimes gives $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ . These are indeterminate forms: they tell you nothing about the limit, which could be any number, or could diverge.

The key standard result is $\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1 \quad (\theta \text{ in radians}).$ Forms like $0 \times \infty$ (for example $x\ln x$ as $x \to 0^{+}$ ) must first be rewritten as a quotient: $x\ln x = \dfrac{\ln x}{1/x}$ .

Key termsindeterminate form

Common mistake

$\frac{0}{0}$ is not $0$ , and $\frac{\infty}{\infty}$ is not $1$ . Always apply a method.

Section 2

l'Hôpital's rule

If $\lim_{x\to a}\dfrac{f(x)}{g(x)}$ is of the form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ , then $\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)},$ provided the right-hand limit exists. The same holds for $x \to \infty$ .

Example: $\lim_{x\to0}\dfrac{e^{3x} - 1}{\sin 2x} = \lim_{x\to0}\dfrac{3e^{3x}}{2\cos 2x} = \dfrac{3}{2}$ .

Key termsl'Hôpital's rule

Common mistake

Using the quotient rule. l'Hôpital's rule differentiates the numerator and the denominator separately.

Common mistake

Applying the rule when the form is not indeterminate, e.g. to $\dfrac{x + 1}{x + 2}$ at $x = 0$ , which is simply $\dfrac{1}{2}$ .

Section 3

Repeated use of the rule

If $\dfrac{f'(x)}{g'(x)}$ is still indeterminate, apply the rule again — and check the form each time.

$\lim_{x\to0}\frac{x - \sin x}{x^{3}} = \lim_{x\to0}\frac{1 - \cos x}{3x^{2}} = \lim_{x\to0}\frac{\sin x}{6x} = \lim_{x\to0}\frac{\cos x}{6} = \frac{1}{6}.$ Each of the first three quotients is $\frac{0}{0}$ at $x = 0$ ; the last is not, so you stop and substitute.

Key termsrepeated application

Exam tip

In an exam, write ' $\frac{0}{0}$ ' (or ' $\frac{\infty}{\infty}$ ') next to every quotient before you differentiate. It earns the reasoning mark.

Section 4

Limits as x tends to infinity

For $\dfrac{\infty}{\infty}$ forms as $x \to \infty$ , the rule compares rates of growth:

$\lim_{x\to\infty}\dfrac{x^{2}}{e^{x}} = \lim\dfrac{2x}{e^{x}} = \lim\dfrac{2}{e^{x}} = 0$ : exponentials beat powers.
$\lim_{x\to\infty}\dfrac{\ln x}{x} = \lim\dfrac{1/x}{1} = 0$ : powers beat logarithms.
$\lim_{x\to\infty}\dfrac{5x^{2} - 3x}{2x^{2} + \ln x} = \dfrac{5}{2}$ : the leading terms decide.

In context, such limits describe long-term behaviour, e.g. $\lim_{n\to\infty}\left(1 + \dfrac{r}{n}\right)^{n} = e^{r}$ is continuous compounding (take logarithms first, then use the rule on $\dfrac{\ln(1 + r/n)}{1/n}$ ).

Key termsrate of growth

Section 5

Using Maclaurin series

Replacing functions by their Maclaurin series turns a limit into algebra: $\frac{x - \sin x}{x^{3}} = \frac{x - \left(x - \frac{x^{3}}{6} + \cdots\right)}{x^{3}} = \frac{1}{6} - \frac{x^{2}}{120} + \cdots \to \frac{1}{6}.$ Series are especially useful when an unknown constant is involved. If $\lim_{x\to0}\dfrac{e^{ax} - 1 - 2x}{x^{2}}$ is finite, the $x$ term of the numerator, $(a - 2)x$ , must vanish, so $a = 2$ and the limit is $\dfrac{a^{2}}{2} = 2$ .

Key termsMaclaurin series

Exam tip

Expand to one power beyond the power in the denominator so you can see the limiting term.

Must know

$\frac{0}{0}$ and $\frac{\infty}{\infty}$ are indeterminate; $\lim_{\theta\to0}\frac{\sin\theta}{\theta} = 1$ .
l'Hôpital: $\lim\frac{f}{g} = \lim\frac{f'}{g'}$ for these forms only. Differentiate top and bottom separately.
Re-check the form before every repeated application.
Rewrite products such as $x\ln x$ as quotients first.
Maclaurin series give an alternative method and handle unknown constants neatly.

That's the notes covered.

Carry on to the next subtopic.