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Date	May 2019	Marks available	9	Reference code	19M.3.AHL.TZ0.Hca_4
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 0
Command term	Find	Question number	Hca_4	Adapted from	N/A

Question

Using L’Hôpital’s rule, find $\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{tan}}\,3x - 3\,{\text{tan}}\,x}}{{{\text{sin}}\,3x - 3\,{\text{sin}}\,x}}} \right)$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{tan}}\,3x - 3\,{\text{tan}}\,x}}{{{\text{sin}}\,3x - 3\,{\text{sin}}\,x}}} \right)$

$\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{3}}\,{\text{se}}{{\text{c}}^2}\,3x - 3\,{\text{se}}{{\text{c}}^2}\,x}}{{{\text{3}}\,{\text{cos}}\,3x - 3\,{\text{cos}}\,x}}} \right)\,\,\,\,\,\left( { = \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{se}}{{\text{c}}^2}\,3x - {\text{se}}{{\text{c}}^2}\,x}}{{{\text{cos}}\,3x - {\text{cos}}\,x}}} \right)\,} \right)$ M1A1A1

Note: Award M1 for attempt at differentiation using l'Hopital's rule, A1 for numerator, A1 for denominator.

METHOD 1

using l’Hopital’s rule again

$= \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{18}}\,{\text{se}}{{\text{c}}^2}\,3x\,{\text{tan}}\,3x - 6\,{\text{se}}{{\text{c}}^2}\,x\,{\text{tan}}\,x}}{{ - 9\,{\text{sin}}\,3x + 3\,{\text{sin}}\,x}}} \right)\,\,\left( { = \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{6}}\,{\text{se}}{{\text{c}}^2}\,3x\,{\text{tan}}\,3x - 2\,{\text{se}}{{\text{c}}^2}\,x\,{\text{tan}}\,x}}{{ - 3\,{\text{sin}}\,3x + {\text{sin}}\,x}}} \right)} \right)$ A1A1

EITHER

$= \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{108}}\,{\text{se}}{{\text{c}}^2}\,3x\,{\text{ta}}{{\text{n}}^2}\,3x + 54\,{\text{se}}{{\text{c}}^4}\,3x - 12\,{\text{se}}{{\text{c}}^2}\,x\,{\text{ta}}{{\text{n}}^2}\,x - 6\,{\text{se}}{{\text{c}}^4}\,x}}{{{\text{ - 27}}\,{\text{cos}}\,3x + 3\,{\text{cos}}\,x}}} \right)$ A1A1

Note: Not all terms in numerator need to be written in final fraction. Award A1 for $54\,{\text{se}}{{\text{c}}^4}\,3x + \ldots - 6\,{\text{se}}{{\text{c}}^4}\,x \ldots -$ . However, if the terms are written, they
must be correct to award A1.

attempt to substitute $x = 0$ M1

$= \frac{{48}}{{ - 24}}$

$\frac{{\text{d}}}{{{\text{d}}x}}\left( {{\text{18}}\,{\text{se}}{{\text{c}}^2}\,3x\,{\text{tan}}\,3x - 6\,{\text{se}}{{\text{c}}^2}\,x\,{\text{tan}}\,x} \right)\left| {_{x = 0} = 48} \right.$ (M1)A1

$\frac{{\text{d}}}{{{\text{d}}x}}\left( { - 9\,{\text{sin}}\,3x + 3\,{\text{sin}}\,x} \right)\left| {_{x = 0} = - 24} \right.$ A1

THEN

$\left( {\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{tan}}\,3x - 3\,{\text{tan}}\,x}}{{{\text{sin}}\,3x - 3\,{\text{sin}}\,x}}} \right)} \right) = - 2$ A1

METHOD 2

$= \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{\frac{3}{{{\text{co}}{{\text{s}}^2}\,3x}} - \frac{3}{{{\text{co}}{{\text{s}}^2}\,x}}}}{{{\text{3}}\,{\text{cos}}\,3x - 3\,{\text{cos}}\,x}}} \right)$ M1

$= \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{co}}{{\text{s}}^2}\,x - {\text{co}}{{\text{s}}^2}\,3x}}{{{\text{co}}{{\text{s}}^2}\,3x\,{\text{co}}{{\text{s}}^2}\,x\left( {{\text{cos}}\,3x - {\text{cos}}\,x} \right)}}} \right)$ A1

$= \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{cos}}\,x + {\text{cos}}\,3x}}{{ - {\text{co}}{{\text{s}}^2}\,3x\,{\text{co}}{{\text{s}}^2}\,x}}} \right)$ M1A1

attempt to substitute $x = 0$ M1

$= \frac{2}{{ - 1}}$

$= - 2$ A1

[9 marks]

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Topic 5 —Calculus » AHL 5.13—Limits and L’Hopitals

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