Stand Up & Math · Teacher Answers

Limits of Trig Expressions
using Maclaurin Series

Not for distribution · HL 5.13 · Full worked solutions + marking guidance
🔐 Student Activity Unlock Codes (case-insensitive)
Phase 1 → Phase 2
MACLRN
Phase 2 → Phase 3
DIVERG
Phase 3 → Phase 4
FINITE
Phase 4 → Complete
TAYLOR
🎨 Oral Launch — Read Aloud to Class

"You've seen limits before. Today you're going to build a tool — Maclaurin series — that can crack limits that your previous techniques can't touch. The warm-up is going to show you exactly why you need something new."

Give code MACLRN immediately at the start — Phase 1 is the warm-up and all groups should begin simultaneously. Only gate DIVERG after Task C is complete on all boards.

1

Warm-up Tasks

Give code MACLRN at the start. Gate DIVERG only after Task C.

Task AA familiar limit — \(\lim_{x\to 0}\dfrac{\sin 5x}{x}\)
\[\lim_{x \to 0} \frac{\sin 5x}{x} = 5 \cdot \lim_{x \to 0} \frac{\sin 5x}{5x} = 5 \cdot 1 = 5\]
Answer: 5

The standard result is \(\displaystyle\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1\). Multiplying top and bottom by 5 reduces this to the standard form.

Most groups will get this immediately. Its purpose is to distinguish the easy case from Task B — the contrast is the pedagogical hook.
Task BA harder denominator — \(\lim_{x\to 0}\dfrac{\sin 3x}{x^3}\)

Direct substitution gives \(\dfrac{0}{0}\) — indeterminate. Attempting the standard trick:

\[\frac{\sin 3x}{x^3} = \frac{3}{x^2} \cdot \frac{\sin 3x}{3x} \to \frac{3}{x^2} \cdot 1 = \frac{3}{x^2} \to \infty\]
The limit diverges to \(+\infty\). It does not exist as a finite number.
This is the motivating problem. Push groups: "So if we can't evaluate it directly, and L'Hôpital gives \(\infty\), what do we need?" The answer is Maclaurin — allow Task C to deliver this.
Task CBridge — for what \(n\) does \(\lim_{x\to 0}\!\left(\dfrac{\sin 3x}{x^3}+\dfrac{n}{x^2}\right)\) exist finitely?

Expand \(\sin 3x\) using the Maclaurin series:

\[\sin 3x = 3x - \frac{(3x)^3}{3!} + \frac{(3x)^5}{5!} - \cdots = 3x - \frac{9x^3}{2} + O(x^5)\] \[\frac{\sin 3x}{x^3} = \frac{3}{x^2} - \frac{9}{2} + O(x^2)\] \[\frac{\sin 3x}{x^3} + \frac{n}{x^2} = \frac{3+n}{x^2} - \frac{9}{2} + O(x^2)\]

For a finite limit as \(x\to 0\), the \(x^{-2}\) term must vanish:

\[3 + n = 0 \implies n = -3\]
\(n = -3\), giving \(\displaystyle\lim_{x\to 0} = -\frac{9}{2}\)
Key insight

The Maclaurin series turns the indeterminate expression into a power series whose leading terms we can read off — making the condition for a finite limit transparent.

Unlock code: MACLRN. This task should be done before giving MACLRN — Task C IS the discovery. Give the code after all boards show \(n=-3\) with a Maclaurin argument.
2

Part (a) — Show that \(n = -3\) is necessary

Give DIVERG once all groups have a complete argument for uniqueness.

Part (a) Show that a finite limit requires \(n = -3\) 3 marks
\[\sin 3x = 3x - \frac{9x^3}{2} + O(x^5)\] \[\frac{\sin 3x}{x^3} = \frac{3}{x^2} - \frac{9}{2} + O(x^2)\] \[f(x) = m + \frac{3}{x^2} - \frac{9}{2} + O(x^2) + \frac{n}{x^2} = m + \frac{3+n}{x^2} - \frac{9}{2} + O(x^2)\] \[\lim_{x\to 0} f(x) \text{ finite} \iff 3 + n = 0 \iff \boxed{n = -3} \qquad \square\]
Common error: Students show that \(n=-3\) works without arguing uniqueness. The word "only" demands this. Prompt: "How do you know no other value could give a finite limit?" The \(x^{-2}\) term grows without bound for any \(n \neq -3\). Unlock code: DIVERG
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Part (b) — Find \(m\)

Give FINITE once all groups have the correct value of \(m\) with full working.

Part (b) Find \(m\) such that \(\displaystyle\lim_{x\to 0} f(x) = 5\) 5 marks

With \(n = -3\):

\[f(x) = m + \frac{\sin 3x}{x^3} - \frac{3}{x^2}\] \[\frac{\sin 3x}{x^3} - \frac{3}{x^2} = \frac{\sin 3x - 3x}{x^3}\] \[\sin 3x - 3x = \left(3x - \frac{9x^3}{2} + O(x^5)\right) - 3x = -\frac{9x^3}{2} + O(x^5)\] \[\frac{\sin 3x - 3x}{x^3} = -\frac{9}{2} + O(x^2) \;\longrightarrow\; -\frac{9}{2} \text{ as } x \to 0\] \[\lim_{x \to 0} f(x) = m + \left(-\frac{9}{2}\right) = m - \frac{9}{2} = 5\] \[\boxed{m = 5 + \frac{9}{2} = \frac{19}{2}}\]
\(m = \dfrac{19}{2}\)

L'Hôpital alternative (also full marks if steps are shown):

\[\frac{\sin 3x - 3x}{x^3} \xrightarrow{\,\text{L'H}\,} \frac{3\cos 3x - 3}{3x^2} \xrightarrow{\,\text{L'H}\,} \frac{-9\sin 3x}{6x} \xrightarrow{\,\text{L'H}\,} \frac{-27\cos 3x}{6} \to -\frac{27}{6} = -\frac{9}{2}\]

Three applications required. Both methods give \(m = \dfrac{19}{2}\).

Students often forget to first combine \(\dfrac{\sin 3x}{x^3} - \dfrac{3}{x^2} = \dfrac{\sin 3x - 3x}{x^3}\) before applying L'Hôpital. Without this step, L'Hôpital is not applicable. Unlock code: FINITE
4

Extension Answers

Give TAYLOR once groups have completed at least one extension task.

Extension 1Replace \(\sin 3x\) with \(\sin 5x\)
\[\sin 5x = 5x - \frac{25x^3 \cdot 5}{6} \cdots = 5x - \frac{125x^3}{6} + O(x^5)\] \[\text{Wait — correct expansion: }\sin 5x = 5x - \frac{(5x)^3}{6} + \cdots = 5x - \frac{125x^3}{6} + O(x^5)\] \[\frac{\sin 5x}{x^3} = \frac{5}{x^2} - \frac{125}{6} + O(x^2)\]

For \(\dfrac{\sin 5x}{x^3} + \dfrac{n}{x^2}\) to have a finite limit: \(5 + n = 0 \implies n = -5\).

\[\lim_{x\to 0}\left(m + \frac{\sin 5x - 5x}{x^3}\right) = m - \frac{125}{6} = 5 \implies m = 5 + \frac{125}{6} = \frac{155}{6}\]
\(n = -5\), \(\;m = \dfrac{155}{6}\)
Note: \((5x)^3/3! = 125x^3/6\), so \(\sin 5x/x^3 \to -125/6\), not \(-9/2\). The structure is identical — only the numbers change.
Extension 2L'Hôpital's route — how many applications?

Three applications of L'Hôpital are needed to reduce \(\dfrac{\sin 3x - 3x}{x^3}\) to a determinate form. Each application handles one power of \(x\) in the denominator.

Efficiency comparison

Maclaurin: write out the first two terms of \(\sin 3x\), cancel, read off the answer — 3 lines of work. L'Hôpital: three separate applications, each requiring a derivative — 8–10 lines of work. For higher-order denominators, Maclaurin scales far better.

Extension 3Full generalisation — \(\sin ax\)
\[\sin ax = ax - \frac{(ax)^3}{6} + O(x^5) = ax - \frac{a^3 x^3}{6} + O(x^5)\] \[\frac{\sin ax}{x^3} = \frac{a}{x^2} - \frac{a^3}{6} + O(x^2)\] \[\text{Finite limit requires: }a + n = 0 \implies n = -a\] \[\lim_{x\to 0} f(x) = m - \frac{a^3}{6}\]
\(n = -a\)  ·  \(\displaystyle\lim_{x\to 0} f(x) = m - \dfrac{a^3}{6}\)

For the limit to equal 5: \(m = 5 + \dfrac{a^3}{6}\). Check with \(a=3\): \(m = 5 + \dfrac{27}{6} = 5 + \dfrac{9}{2} = \dfrac{19}{2}\) ✓

Unlock code: TAYLOR
📋 Consolidation — select 2–3 boards to debrief
  • Task C: How did you know only one value of \(n\) works? What makes the \(x^{-2}\) term special?
  • Part (a): What does "show that" demand beyond just substituting \(n=-3\)?
  • Part (b): Why must you combine the fractions before applying L'Hôpital?
  • Extension 3: What does the formula \(m = 5 + a^3/6\) tell us geometrically about the function?