Peskin and Schroeder spinor high-energy limit (5.26 and A.20) P&S say the high-energy limit of spinor $u^s (p)$ is 
$ \sqrt{2E}  {1 \over 2}    (1-\widehat{p} . {\sigma}) \xi^s   $ and similar for the right-handed spinor (formulae 5.26 and A.20). I can't seem to derive this. How do you get this from $\sqrt{p . \sigma } \xi^s    $?
 A: This is trivial to see using the explicit (and the generically more useful) form of the square root,
$$
\sqrt{p.\sigma} \equiv \frac{E_p+m-{\bf \sigma}.\bf{p}}{\sqrt{2(E_p+m)}}
$$
Then we just send $m\rightarrow 0, {\bf p}\rightarrow E_p\hat{p}$ and we immediately get the term you write above,
$$
\sqrt{p.\sigma} \equiv \frac{E_p-E_p{\bf \sigma}.\hat{p}}{\sqrt{2(E_p)}} = \sqrt{\frac{E_p}{2}}(1-\sigma.\hat{p})
$$
A: A little late to the party, but hopefully this will also be of use to someone else. 
In P&S a short remark can also be found on the calculation of the square root of the matrix on page 46, around equation 3.50, which I assume hints at starting with diagonalising the matrix using the diagonal matrix of eigenvalues and the matrix consisting of eigenvectors, and its inverse.
To compute the square root of the matrix, first diagonalise $p\cdot \sigma$ as
$$p \cdot \sigma = U 
\begin{pmatrix}
\lambda_0 & 0 \\
0 & \lambda_1 \\ 
\end{pmatrix}
U^{-1}
$$
where $U$ is the matrix consisting of eigenvectors. We then have:
$$ \sqrt{p \cdot \sigma} = U 
\begin{pmatrix}
\sqrt{\lambda_0} & 0 \\
0 & \sqrt{\lambda_1} \\ 
\end{pmatrix}
U^{-1}
$$
Using this approach you will find that this gives the result as shown in P&S.
A: Yet another way to go about it... In the relativistic limit,
$$\sqrt{p\cdot \sigma}\equiv\sqrt{E+\mathbf{p}\cdot\boldsymbol\sigma}\approx\sqrt{E+E\hat{p}\cdot\boldsymbol{\sigma}}=\sqrt{2E}\sqrt{\frac{1}{2}+\frac{1}{2}\hat{p}\cdot\boldsymbol\sigma},$$
where $\hat{p}=\frac{\mathbf{p}}{|\mathbf{p}|}\approx\frac{\mathbf{p}}{E}$. Now note that
$$\left(\frac{1}{2}+\frac{1}{2}\hat{p}\cdot\boldsymbol\sigma\right)^2=\frac{1}{4}+\frac{1}{2}\hat{p}\cdot\boldsymbol\sigma+\frac{1}{4}(\hat{p}\cdot\boldsymbol\sigma)^2.$$
Now we just use the fact that
$$(\hat{p}\cdot\boldsymbol\sigma)^2=\sum_i\hat{p}_i^2\sigma_i^2+\sum_{i<j}\hat{p}_i\hat{p}_j\{\sigma_i,\sigma_j\}=\sum_i\hat{p}_i^2=|\hat{p}|=1$$
to get
$$\left(\frac{1}{2}+\frac{1}{2}\hat{p}\cdot\boldsymbol\sigma\right)^2=\frac{1}{2}+\frac{1}{2}\hat{p}\cdot\boldsymbol\sigma.$$
Therefore, $(1/2)(1+\hat{p}\cdot\boldsymbol\sigma)$ is its own square root.
A: Don't know if anyone is still expecting a reply to this, but here is how I managed to find the result. Peskin&Schroeder found result (3.50) by doing the calculation in a specialized frame (boost along z-axis) which resulted in (3.49). If you apply the high energy limit in that same frame by setting E = p3 and rewrite it in a covariant form (using a dot product with unit momentum vector ^p instead of p3), then you get the desired result which is immediately valid in any frame due to the covariant notation.
To anyone who might be concerned...
