= Solution
{c}
At the interface, the tangential <velocity> is $u(x,0)=xf'(0)$ and the normal <velocity> remains zero. The preceding <error function> expression gives
$$
g(0)=f_1'(0)=\frac{\sqrt{2\pi}}4=\sqrt{\frac\pi8}.
$$
Hence, within the leading <boundary layer> model,
$$
\boxed{u(x,0)=x\left[1+\sqrt{\frac\pi8}\,\gamma+O(\gamma^2)\right],
\qquad w(x,0)=0.}
$$
Allowing the formally discarded finite-<Reynolds number> terms gives an additional order-$\delta^2$ error, smaller than the order-$\gamma$ correction under the assumed ordering. The coefficient is $\sqrt{\pi/8}\approx0.626657$.
Thus \b[positive surface shear increases the outward surface speed linearly]: for $x>0$ it increases $u$, and for $x<0$ it makes $u$ more negative. The <stagnation point> remains at $x=0$. A negative coefficient would result from using the printed $+\infty$ condition literally, and would describe the different half-line problem identified in part (ii).
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