A question related to purification theorem I am stuck at some point in part b, I would be very happy if you could help. My calculations are below the question.

a)
P1 plays N with probability p. P2 plays N with probability q.
Mixed Nash: $((\frac{1}{2},N),((\frac{1}{2},N))$
b)
P1 plays N if $(1+x\epsilon_1)q+x\epsilon_1(1-q)&gt;q0+(1-q)1$
$\epsilon_1&gt;\frac{1-2q}{x}$
P2 plays N if $(x\epsilon_2-1)p+x\epsilon_2(1-p)&gt;p0+(1-p)(-1)$
$\epsilon_2&gt;\frac{2p-1}{x}$
The distribution is not necessarily uniform. pdf's are not specified explicitly, so I need to find a general solution.
$p=Pr(\epsilon_1&gt;\frac{1-2q}{x})=1-Pr(\epsilon_1&lt;\frac{1-2q}{x})=1-F_{\epsilon_1} (\epsilon_1&lt;\frac{1-2q}{x})$
$q=Pr(\epsilon_2&gt;\frac{2p-1}{x})=1-Pr(\epsilon_2&lt;\frac{2p-1}{x})=1-F_{\epsilon_2} (\epsilon_2&lt;\frac{2p-1}{x})$
I am stuck at this point.
I have never done such purification exercises, but I would approach it like that. As you state,
$p=Pr(\epsilon_1&gt;\frac{1-2q}{x})=1-Pr(\epsilon_1&lt;\frac{1-2q}{x})=1-F (\frac{1-2q}{x}),$
$q=Pr(\epsilon_2&gt;\frac{2p-1}{x})=1-Pr(\epsilon_2&lt;\frac{2p-1}{x})=1-G (\frac{2p-1}{x}),$
where $F$ and $G$ are the cdfs corresponding to densities $f=F'$ and $g=G'$. Because they are cdfs, they are weakly increasing. For simplicity, suppose they are strictly increasing so that their inverses clearly exist (i.e., I assume full support. If there were holes, you could still define appropriately some inverse for the flat parts).
Then, combine those two equations to
$1+x F^{-1}(1-p) - 2 G (\frac{2p-1}{x})=0$ (and another similar formulation for q). The LHS of this formula decreasing in $p$. Note that for $p=0$, we have $1+xF^{-1}(1)-2G(\frac{-1}{x})=1+x-0&gt;0$ on the LHS such that the equation never holds, $p$ must be larger than zero. For $p=1$, have on the LHS $1+xF^{-1}(0)-2G(1/x)=1-2G(1/x)$ which can be more or less than 0. That is, for some $x$ there is a pure strategy equilibrium with $p=1$ and for others there is a mixed strategy equilibrium with $p&lt;1$.
Now consider the same equation that solves for the mixed strategy and plug in $p=\frac{1-x}{2}$ such that the LHS above is $1+xF^{-1}(\frac{1+x}{2})- 2 G(-1)=1+x F^{-1}(\frac{1+x}{2})- 0&gt;0$ for small $x$, meaning we have to increase $p$ to fit the equilibrium equation LHS=0.
Next consider the same equation that solves for the mixed strategy and plug in $p=\frac{1+x}{2}$ such that the LHS above is $1+xF^{-1}(\frac{1-x}{2})- 2 G(1)=x F^{-1}(\frac{1+x}{2})-1&lt;0$ for small $x$, meaning we have to decrease $p$ to fit the equilibrium equation LHS=0.
Thus, for small $x$, the equilibrium $p$ is such that $p\in [\frac{1-x}{2},\frac{1+x}{2}]$, leaving only candidate $p\to \frac{1}{2}$ as $x\to0$.