The standard generalised extreme value (GEV) distribution is given by $H_{\xi}$ which is
$exp(-(1+\xi x)^{-1/\xi}$ if $\xi<>0$ and
$exp(-e^{-x})$ if $\xi=0$
In the lecture notes it is stated
$1-H_\xi (x)$ approximatle equals $e^{-x}$ for $\xi=0$ for $x_{H_\xi}$ going to infinity which is the Gumbel distribution.
$1-H_\xi (x)$ approximatle equals $(\xi x)^{-1/\xi}$ for $\xi=0$ for $x_{H_\xi}$ going to infinity which is the Fréchet distribution.
I would like to do the math to derive the Gumbel and the Fréchet from the GEV to understand deriving limits better (I seem to have some deficits). I would be grateful for a solution or a textbook hint.
Many thanks.
To restate your question, the form of the generalized extreme value cumulative distribution function is given for the cases $\xi\not=0$ and $\xi=0$. The case $\xi=0$ is known as the Gumbel distribution. You are asking for a demonstration of how the formula for the $\xi=0$ cases arises as a limit of the $\xi\not=0$ case.
By inspection, this amounts to the assertion $$ \lim_{\xi\to0} (1+\xi x)^{-1/\xi}=e^{-x}. $$ Taking negative $\log$ of both sides, it is equivalent to $$ \lim_{\xi\to0} \frac{\log(1+\xi x)}{\xi}=x.\qquad(\star) $$ The left side is recognized as the partial derivative of $\log(1+\xi x)$ with respect to $\xi$, evaluated at $\xi=0$. Using basic rules of differentiation, we have $$ \frac{\partial }{\partial \xi}\log(1+\xi x)=\frac{x}{1+\xi x}. $$ Therefore $$ \lim_{\xi\to0} \frac{\log(1+\xi x)}{\xi}=\left.\frac{\partial }{\partial \xi}\log(1+\xi x)\right|_{\xi=0}=x, $$ establishing $(\star)$ and therefore proving that the Gumbel CDF is the $\xi=0$ limit of the GEV CDF.