32. For a random variable $\xi$, having a $\mathscr{N}(0,1)$-distribution,

$$
\mathrm{P}(\xi \geqslant x) \sim \frac{\varphi(x)}{x}, \quad x \rightarrow \infty, \quad \text { where } \varphi(x)=\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}} .
$$

Find the corresponding asymptotic for a random variable $\zeta$, having a gamma-distribution with density

$$
f(x)=\frac{I(x>0)}{\Gamma(\alpha)} \cdot x^{\alpha-1} e^{-x}, \quad \alpha>0
$$
Solution. The desired asymptotic is

$$
\mathrm{P}(\zeta \geqslant x)=\frac{1}{\Gamma(\alpha)} \int_{x}^{\infty} z^{\alpha-1} e^{-z} d z \sim f(x)=\frac{x^{\alpha-1} e^{-x}}{\Gamma(\alpha)}, \quad x \rightarrow \infty
$$

Indeed, by L'Hôpital's rule

$$
\begin{aligned}
\lim _{x \rightarrow \infty} \frac{\int_{x}^{\infty} z^{\alpha-1} e^{-z} d z}{x^{\alpha-1} e^{-x}}=\lim _{x \rightarrow \infty} & \frac{\left(\int_{x}^{\infty} z^{\alpha-1} e^{-z} d z\right)^{\prime}}{\left(x^{\alpha-1} e^{-x}\right)^{\prime}}= \\
& =\lim _{x \rightarrow \infty} \frac{-x^{\alpha-1} e^{-x}}{-x^{\alpha-1} e^{-x}\left[1+(\alpha-1) x^{-1}\right]}=1
\end{aligned}
$$