MathAA — Print Cards
Limits of Trig Expressions · topic5-calculus/trig-limits
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Phase 1: MACLRN
Phase 2: DIVERG
Phase 3: FINITE
Phase 4: TAYLOR

BTC · AHL 5.13
Limits of Trig Expressions
8 marks total · HL
\[f(x) = m + \frac{\sin 3x}{x^3} + \frac{n}{x^2}, \qquad m,\, n \in \mathbb{R}\]
⏱ ~25 min👥 Groups of 3 · VNPS📐 Topic 5 · Calculus

Warm-up · Task A
Familiar limit

Try substituting small values of \(x\), then confirm with the standard result.

\[\lim_{x \to 0} \frac{\sin 5x}{x} = \;?\]
What rule did you use?
Warm-up · Task B
Harder denominator

What indeterminate form do you get? What technique can you use?

\[\lim_{x \to 0} \frac{\sin 3x}{x^3} = \;?\]
Does this limit exist?

Warm-up · Task C — bridge
When can two infinities cancel out?

For what value of \(n\) does the limit below exist as a finite number? Show your full reasoning — explain why only one value works.

\[\lim_{x \to 0} \left(\frac{\sin 3x}{x^3} + \frac{n}{x^2}\right)\]
Hint: expand \(\sin 3x\) as a Maclaurin series and divide by \(x^3\).

Main task · Part (a)
Show that \(n = -3\) is the only possible value
3 marks
\[f(x) = m + \frac{\sin 3x}{x^3} + \frac{n}{x^2}, \qquad m,\,n \in \mathbb{R}\]

Show that a finite limit for \(\displaystyle\lim_{x \to 0} f(x)\) can only exist when \(n = -3\).


Main task · Part (b)
Find \(m\)
5 marks

Using \(n = -3\) from part (a), find the value of \(m\) for which:

\[\lim_{x \to 0} f(x) = 5\]

Show every step of the calculation. Exact form required.


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

What changes? Find the new \(n\) and the new \(m\) if the limit still equals 5.

Extension 2
L'Hôpital's route

Solve part (b) using L'Hôpital's rule. How many applications? Which method is more efficient?

Extension 3 — generalise
General case
\[f(x) = m + \frac{\sin ax}{x^3} + \frac{n}{x^2}, \qquad a \in \mathbb{R}\]
Find \(n\) in terms of \(a\). Find \(\displaystyle\lim_{x\to 0} f(x)\) in terms of \(m\) and \(a\). If the limit equals \(L\), what is \(m\)?