18N.3.AHL.TZ0.HCA_2

pestleMathematicsAAHLPaper 318N· ahl-5-13-limits-and-lhopitalssource ↗

Use L’Hôpital’s rule to determine the value of

$\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{{\text{e}}^{ - 3x}}^{^2} + 3\,{\text{cos}}\left( {2x} \right) - 4}}{{3{x^2}}}} \right)$

[5]

Hence find $\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{\int_0^x {\left( {{{\text{e}}^{ - 3t}}^{^2} + 3\,{\text{cos}}\left( {2t} \right) - 4} \right)} \,{\text{d}}t}}{{\int_0^x {3{t^2}} \,{\text{d}}t}}} \right)$ .

[3]

Markscheme / solution

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

$\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{{{\text{e}}^{ - 3x}}^{^2} + 3\,{\text{cos}}\,2x - 4}}{{3{x^2}}} = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)$

$= \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{ - {\text{6}}x{{\text{e}}^{ - 3x}}^{^2} - 6\,{\text{sin}}\,2x}}{{6x}} = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)$ M1A1A1

$= \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{ - {\text{6}}{{\text{e}}^{ - 3x}}^{^2} + 36{x^2}{{\text{e}}^{ - 3x}}^{^2} - 12\,{\text{cos}}\,2x}}{6}$ A1

= −3 A1

[5 marks]

$\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{\int_0^x {\left( {{{\text{e}}^{ - 3t}}^{^2} + 3\,{\text{cos}}\,2t - 4} \right)} \,{\text{d}}t}}{{\int_0^x {3{t^2}} \,{\text{d}}t}}} \right)$ is of the form $\frac{0}{0}$

applying l’Hôpital´s rule (M1)

$\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{{{\text{e}}^{ - 3x}}^{^2} + 3\,{\text{cos}}\,2x - 4}}{{3{x^2}}}$ (A1)

= −3 A1

[3 marks]

Examiners’ report

[N/A]