Using l’Hôpital’s rule, show that

$\mathop {\lim }\limits_{x \to \infty } {x^n}{{\text{e}}^{ - x}} = 0$ where $n \in \mathbb{N}$.

Show that, for $n \in {\mathbb{Z}^ + }$,

$${I_n} = \alpha {{\text{e}}^{ - 1}} + \beta n{I_{n - 1}}$$

where $\alpha$, $\beta$ are constants to be determined.

Determine the value of ${I_3}$, giving your answer as a multiple of ${{\text{e}}^{ - 1}}$.

consider $\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{{\text{e}}^x}}}$     M1

its value is $\frac{\infty }{\infty }$ so we use l’Hôpital’s rule

$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{{\text{e}}^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n{x^{n - 1}}}}{{{{\text{e}}^x}}}$     (A1)

its value is still $\frac{\infty }{\infty }$ so we need to differentiate numerator and denominator a further $n - 1$ times     (R1)

this gives $\mathop {\lim }\limits_{x \to \infty } \frac{{n!}}{{{{\text{e}}^x}}}$     A1

since the numerator is finite and the denominator $\to \infty$, the limit is zero     AG

[4 marks]

attempt at integration by parts $\left( {{I_n} =  - \int_1^\infty  {{x^n}{\text{d}}({{\text{e}}^{ - x}})} } \right)$     M1

${I_n} =  - [{x^n}{{\text{e}}^{ - x}}]_1^\infty  + n\int_1^\infty  {{x^{n - 1}}{{\text{e}}^{ - x}}{\text{d}}x}$     A1A1

$= {{\text{e}}^{ - 1}} + n{I_{n - 1}}$     A1

$\alpha  = \beta  = 1$

[9 marks]

${I_3} = {{\text{e}}^{ - 1}} + 3{I_2}$     M1

$= {{\text{e}}^{ - 1}} + 3({{\text{e}}^{ - 1}} + 2{I_1})$     A1

$= 4{{\text{e}}^{ - 1}} + 6({{\text{e}}^{ - 1}} + {I_0})$     A1

$= 4{{\text{e}}^{ - 1}} + 6{{\text{e}}^{ - 1}} + 6\int_1^\infty  {{{\text{e}}^{ - x}}{\text{d}}x}$

$= 10{{\text{e}}^{ - 1}} - 6[{{\text{e}}^{ - x}}]_1^\infty$     A1

$= 16{{\text{e}}^{ - 1}}$     A1

[9 marks]