Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) §6.2 The Exponential e, #59 · Calculus · Differentiation rules

This exercise shows that … as

This exercise shows that $F (x) = x^{n} / e^{x} \to 0$ as $x \to \infty$ .
(a) Find $d F / d x$ . Notice that $F (x)$ decreases for $x > n > 0$ .
The maximum of $x^{n} / e^{x}$ , at $x = n$ , is $n^{n} / e^{n}$ .
(b) $F (2 x) = (2 x)^{n} / e^{2 x} = 2^{n} x^{n} / e^{x} \cdot e^{- x} ⩽ 2^{n} n^{n} / e^{n} \cdot e^{- x}$ .
Deduce that $F (x) \to 0$ as $x \to \infty$ . Thus $F (x) \to 0$ .

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Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §6.2 The Exponential e, #59

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