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]