Answer
$0$
Work Step by Step
$$\lim\limits_{x \to \infty}x^{5}e^{-0.001x}=\lim\limits_{x \to \infty}\frac{x^{5}}{e^{0.001x}}=\frac{\infty}{\infty}$$
Using the l'Hospital's rule it follows:
$$\lim\limits_{x \to \infty}\frac{x^{5}}{e^{0.001x}}=\lim\limits_{x \to \infty}\frac{5x^{4}}{0.001e^{0.001x}}=\frac{\infty}{\infty}$$
Using the l'Hospital's rule for the second time it follows:
$$\lim\limits_{x \to \infty}\frac{5x^{4}}{0.001e^{0.001x}}=\lim\limits_{x \to \infty} \frac{20x^{3}}{(0.001)^{2}e^{0.001x}}=\frac{\infty}{\infty}$$
Using the l'Hospital's rule for the third time it follows:
$$\lim\limits_{x \to \infty} \frac{20x^{3}}{(0.001)^{2}e^{0.001x}}=\lim\limits_{x \to \infty} \frac{60x^{2}}{(0.001)^{3}e^{0.001x}}=\frac{\infty}{\infty}$$
Using the l'Hospital's rule for the fourth time it follows:
$$\lim\limits_{x \to \infty} \frac{60x^{2}}{(0.001)^{3}e^{0.001x}}=\lim\limits_{x \to \infty} \frac{120x}{(0.001)^{4}e^{0.001x}}=\frac{\infty}{\infty}$$
Using the l'Hospital's rule for the fifth time it follows:
$$\lim\limits_{x \to \infty} \frac{120x}{(0.001)^{4}e^{0.001x}}=\lim\limits_{x \to \infty} \frac{120}{(0.001)^{5}e^{0.001x}}=0$$
