calculating a limit with parameter I want to determine the limit
\begin{equation}
\frac{2p - 1}{\sqrt{1 - 4p(1-p)z^{(2p-1)^{2}}}}
\end{equation}
for $p \searrow \frac{1}{2}$, $p \in [0,1]$. 
I have make lot of effort, differentiate again and again using l Hospital rule. But I don't come to a solution. Can anyone help?
 A: Since you seems to care about the real domain, there is a definition issue for $z<0$, so in the following I will only consider $z\in]0,1[$.
Note: for $z=0$ we have $f(p,z)=2p-1\to 0$ trivially.
Let's substitute $p=\frac 12+u$ with $u\to 0$.
$f(u,z)=\dfrac{2u}{\sqrt{1-z^{4u^2}+4u^2z^{4u^2}}}=2u\exp\left(-\frac 12\ln(g(u,z))\right)$ 
Where 
$\require{cancel}\begin{array}{ll}g(u,z)
&=1-z^{4u^2}+4u^2z^{4u^2}\\
&=1-\exp(4u^2\ln(z))+4u^2\exp(4u^2\ln(z))\\
&=1-(1+4u^2\ln(z)+o(u^2))+4u^2(1+\cancel{4u^2\ln(z)}+o(u^2))\\
&=4u^2(1-\ln(z))+o(u^2)
\end{array}$
Thus 
$\begin{array}{ll}f(u,z)
&=2u\exp\left(-\frac 12\ln(4u^2)-\frac 12\ln(1-\ln(z)+o(1))\right)\\
&=\dfrac{2u}{2|u|}\times\dfrac 1{\sqrt{1-\ln(z)+o(1)}}\\
&\sim\dfrac{\operatorname{sgn}(u)}{\sqrt{1-\ln(z)}}\end{array}$
Unfortunately we have a sign issue for the convergence, the limit changes sign in $\frac 12^+$ and $\frac 12^-$, but there is a left limit and a right limit.
You can probably fix that by taking the absolute value of the numerator (namely $|2p-1|$) and have a limit $\dfrac 1{\sqrt{1-\ln(z)}}$.
