My statistics teacher told us the following asymptotic result: $X \sim N(0,1) $ $$ P(X > u) \underset{u \rightarrow +\infty}{\sim} \frac{1}{u} \exp\left(-\frac{u^2}{2}\right). $$
Do you know how to demonstrate this (or if there is a mistake?) Thank you.
The expression you gave is close (you missed the factor $\frac{1}{\sqrt{2\pi}}$) to the correct asymptotic behavior of the survival function of standard normal distribution:
\begin{align*}
1 - \Phi(x) \sim \frac{1}{x}\varphi(x), \tag{1}
\end{align*}
where $\Phi$ and $\varphi$ are CDF and PDF of standard normal random variable respectively.
$(1)$ is a corollary of the inequality (for fixed $x > 0$) \begin{align*} (x^{-1} - x^{-3})\varphi(x) < 1 - \Phi(x) < x^{-1}\varphi(x). \tag{2} \end{align*}
To prove $(2)$, notice the obvious inequality: \begin{align*} (1 - 3t^{-4})\varphi(t) < \varphi(t) < (1 + t^{-2})\varphi(t), \; t > x. \tag{3} \end{align*} Integrating three expressions above from $x$ to $\infty$ yields $(2$) -- note that each term in $(2)$ is the primitive function of each term in $(3)$ times $-1$.
