A2 June 2021 Paper 1 Q9
9 Use l’Hôpital’s rule to show that
\[\lim_{x \to \infty}\left(x\mathrm{e}^{-x}\right) = 0\]Fully justify your answer. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Defines \(\mathrm{f}(x)\) and \(\mathrm{g}(x)\) or uses the correct \(\mathrm{f}(x)\) and \(\mathrm{g}(x)\) with l’Hôpital’s rule | M1 | 3.1a |
| Explains how \(\mathrm{f}(x)\) and \(\mathrm{g}(x)\) fulfil the requirements for l’Hôpital’s rule | E1 | 2.4 |
| Obtains \(\dfrac{\mathrm{f}^{\prime}(x)}{\mathrm{g}^{\prime}(x)} = \mathrm{e}^{-x}\) | A1 | 1.1b |
| Uses correct reasoning to obtain the required limit The explanation of the requirements for l’Hôpital’s rule is not needed for this mark | R1 | 2.1 |
| (4 marks) |
Typical solution
Let \(\mathrm{f}(x) = x,\ \mathrm{g}(x) = \mathrm{e}^x\)
Then \(x\mathrm{e}^{-x} = \dfrac{\mathrm{f}(x)}{\mathrm{g}(x)}\)
and \(\mathrm{f}(x)\) and \(\mathrm{g}(x)\) both tend to \(\infty\) as \(x \to \infty\)
\[\therefore \lim_{x \to \infty}\left(x\mathrm{e}^{-x}\right) = \lim_{x \to \infty}\left(\frac{\mathrm{f}(x)}{\mathrm{g}(x)}\right) = \lim_{x \to \infty}\left(\frac{\mathrm{f}^{\prime}(x)}{\mathrm{g}^{\prime}(x)}\right)\]\(\mathrm{f}^{\prime}(x) = 1\) and \(\mathrm{g}^{\prime}(x) = \mathrm{e}^x\)
\[\therefore \lim_{x \to \infty}\left(x\mathrm{e}^{-x}\right) = \lim_{x \to \infty}\left(\frac{1}{\mathrm{e}^x}\right) = \lim_{x \to \infty}\left(\mathrm{e}^{-x}\right)\]\(= 0\) as required