Moderate -0.5 This is a straightforward application of l'Hôpital's rule requiring students to rewrite xe^{-x} as x/e^x (indeterminate form ∞/∞), differentiate once to get 1/e^x, and conclude the limit is 0. While it's a Further Maths question, it's a direct, single-application problem with no conceptual subtlety—easier than average even for Further Maths standards.
Question 9:
9 | Defines and or uses
the correct and with
l’Hopital’fs( 𝑥𝑥ru)le g(𝑥𝑥)
f(𝑥𝑥) g(𝑥𝑥) | 3.1a | M1 | Let
Then 𝑥𝑥
f(𝑥𝑥) = 𝑥𝑥, g(𝑥𝑥) = e
and − a𝑥𝑥nd f(𝑥𝑥) both tend to
𝑥𝑥e = g(𝑥𝑥)
as
f(𝑥𝑥) g(𝑥𝑥) ∞
𝑥𝑥 → ∞
′
−𝑥𝑥 f(𝑥𝑥) f (𝑥𝑥)
∴ 𝑥𝑥li→m∞(𝑥𝑥e an)d= 𝑥𝑥li→m∞� � = 𝑥𝑥li→m∞� ′ �
g(𝑥𝑥) g (𝑥𝑥)
′ ′ 𝑥𝑥
f (𝑥𝑥) = 1 g (𝑥𝑥) = e
−𝑥𝑥 1 −𝑥𝑥
as required
∴ 𝑥𝑥li→m∞(𝑥𝑥e ) = 𝑥𝑥li→m∞� 𝑥𝑥� = 𝑥𝑥li→m∞(e )
e
= 0
Explains how and
fulfil the requirements for
l’Hôpital’s rulef (𝑥𝑥) g(𝑥𝑥) | 2.4 | E1
Obtains
′
f (𝑥𝑥) −𝑥𝑥
′ | 1.1b | A1
g (𝑥𝑥) = e
Uses correct reasoning to
obtain the required limit
The explanation of the
requirements for l’Hôpital’s
rule is not needed for this
mark | 2.1 | R1
Total | 4
Q | Marking Instructions | AO | Marks | Typical solution
Use l'Hôpital's rule to show that
$$\lim_{x \to \infty} (xe^{-x}) = 0$$
Fully justify your answer.
[4 marks]
\hfill \mbox{\textit{AQA Further Paper 1 2021 Q9 [4]}}