Supremum $\sup_{x\in (0,+\infty)}|F(x,t)|=\sup_{x\in (0,+\infty)}\dfrac{e^{-t^2}}{x+|t|}$ 
Let :
  \begin{aligned}F \colon (0,+\infty)\times \mathbb{R} &\longrightarrow \mathbb{R} \\(x,t) &\longmapsto F(x,t)=\dfrac{e^{-t^2}}{x+|t|}.\end{aligned}



*

*How they do to find  $\displaystyle \sup_{x\in (0,+\infty)}|F(x,t)|$


$$ \fbox{ $\sup_{x\in (0,+\infty)}|F(x,t)|=\sup_{x\in (0,+\infty)}\dfrac{e^{-t^2}}{x+|t|}=\dfrac{e^{-t^2}}{|t|} $} $$
Indeed,
$\forall t\in \mathbb{R}\quad F(.,t)=\dfrac{e^{-t^2}}{x+|t|}$
$$\frac{\partial F}{\partial x}(x,t)=\dfrac{-e^{-t^2}}{(x+|t|)^2} $$
I'm stuck here 

The supremum of this value is achieved when the denominator has the infimum value which is achieved for $x=0$.
