Limits of Trig Expressions — Teacher Answer Booklet

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Warm-up Tasks

Give MACLRN at the start. Give DIVERG only when a group can articulate why only one value of \(n\) works in Task C.

Task ALimit of \(\sin 5x / x\)

\lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} 5 \cdot \frac{\sin 5x}{5x} = 5 \cdot 1 = 5

Using the standard result \(\displaystyle\lim_{\theta \to 0}\dfrac{\sin\theta}{\theta} = 1\) with \(\theta = 5x\).

Task BLimit of \(\sin 3x / x^3\)

Direct substitution gives \(\dfrac{0}{0}\) — indeterminate. Use the Maclaurin series:

\frac{\sin 3x}{x^3} = \frac{3x - \dfrac{27x^3}{6} + \cdots}{x^3} = \frac{3}{x^2} - \frac{9}{2} + O(x^2)

As \(x \to 0\) the \(\dfrac{3}{x^2}\) term dominates — the limit does not exist (diverges to \(+\infty\)).

Pedagogical note Students often expect a finite answer. The divergence here is the key insight that makes Task C meaningful — they need to see why the \(1/x^2\) term matters before the cancellation idea lands.

Task CFinding \(n\) for a finite limit of \(g(x)\)

g(x) = \frac{\sin 3x}{x^3} + \frac{n}{x^2} = \frac{3}{x^2} - \frac{9}{2} + \cdots + \frac{n}{x^2} = \frac{3+n}{x^2} - \frac{9}{2} + \cdots\] \[\text{For a finite limit: } 3 + n = 0 \implies n = -3 \qquad \therefore \lim_{x \to 0} g(x) = -\frac{9}{2}

\(n = -3\) is the unique value. Any other value leaves a \(\dfrac{1}{x^2}\) term that diverges.

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Part (a) — Show \(n = -3\)

Give FINITE only when the argument is complete and explicit on the board. Uniqueness must be argued.

Part (a) Show that a finite limit can only exist when \(n = -3\)

\sin 3x = 3x - \frac{(3x)^3}{3!} + \cdots = 3x - \frac{9x^3}{2} + \cdots\] \[\frac{\sin 3x}{x^3} = \frac{3}{x^2} - \frac{9}{2} + O(x^2)\] \[f(x) = m + \frac{3}{x^2} - \frac{9}{2} + \cdots + \frac{n}{x^2} = m + \frac{3+n}{x^2} - \frac{9}{2} + \cdots\] \[\text{For } \lim_{x \to 0} f(x) \text{ to be finite: } 3 + n = 0 \implies \boxed{n = -3} \qquad \square

Common error Students may show that \(n=-3\) works without arguing uniqueness. The word "only" in the question demands this. Prompt: "How do you know no other value could give a finite limit?"

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Part (b) — Find \(m\)

Give TAYLOR once part (b) is complete.

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

f(x) = m + \frac{\sin 3x}{x^3} - \frac{3}{x^2}\] \[\frac{\sin 3x}{x^3} = \frac{3}{x^2} - \frac{9}{2} + O(x^2)\] \[f(x) = m + \frac{3}{x^2} - \frac{9}{2} + \cdots - \frac{3}{x^2} = m - \frac{9}{2} + O(x^2)\] \[\lim_{x \to 0} f(x) = m - \frac{9}{2} = 5 \implies \boxed{m = \frac{19}{2}}

L'Hôpital alternative Students must first combine: \(\dfrac{\sin 3x}{x^3} - \dfrac{3}{x^2} = \dfrac{\sin 3x - 3x}{x^3}\). This gives \(\dfrac{0}{0}\). Three applications of L'Hôpital yield \(\dfrac{-27\cos 3x}{6} \to -\dfrac{9}{2}\). Full marks if all steps are shown.

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Extension Answers

For fast-finishing groups. Discuss during consolidation.

Extension 1Replace \(\sin 3x\) with \(\sin 5x\)

\sin 5x = 5x - \frac{125x^3}{6} + \cdots \implies \frac{\sin 5x}{x^3} = \frac{5}{x^2} - \frac{125}{6} + \cdots\] \[\text{For finite limit: } n = -5\] \[\lim_{x \to 0} f(x) = m - \frac{125}{6} = 5 \implies m = \frac{155}{6}

Extension 3General case — \(\sin ax\)

\sin ax = ax - \frac{a^3x^3}{6} + \cdots \implies \frac{\sin ax}{x^3} = \frac{a}{x^2} - \frac{a^3}{6} + \cdots\] \[\text{For finite limit: } n = -a\] \[\lim_{x \to 0} f(x) = m - \frac{a^3}{6} = L \implies m = L + \frac{a^3}{6}

Check: \(a=3 \Rightarrow n=-3,\ m = 5 + \dfrac{27}{6} = 5 + \dfrac{9}{2} = \dfrac{19}{2}\) ✓

ConsolidationKey ideas to surface — in order

Dominance near zero: \(\dfrac{1}{x^3} \gg \dfrac{1}{x^2} \gg\) constants as \(x \to 0\). For a finite limit, infinite terms must cancel exactly.
Maclaurin is the key tool: \(\sin 3x = 3x - \dfrac{9x^3}{2} + \cdots\). Dividing by \(x^3\) reveals a hidden \(\dfrac{1}{x^2}\) term.
Uniqueness of \(n\): Only \(n = -3\) cancels the \(\dfrac{1}{x^2}\) term. Any other value leaves a divergent piece.
The limit after cancellation: \(m - \dfrac{9}{2} = 5\), so \(m = \dfrac{19}{2}\).
IB language: "Show that … can only" is a proof of both existence and uniqueness. Students must argue both.

Gallery walk note Reference student boards in consolidation: "I saw three groups use Maclaurin — let's follow that logic together." If any group used L'Hôpital, invite them to walk through step 2 as a contrast.