Let $\frac{1}{3} < \alpha<\frac{1}{2} $; $ \ 0<\epsilon \ll 1$ and assume we have the following inner and outer solutions for an ODE, both valid in the overlap region $\epsilon^{2(1-\alpha)} \lesssim x \lesssim \epsilon^{\frac{2}{3} (1+ \alpha)}$
$$y_{out} (x) = \frac{1}{x^{1/4}} \left[ \cos \left(\frac{\sqrt{x}}{\epsilon} - \frac{\pi}{4}\right) + A \sin \left(\frac{\sqrt{x}}{\epsilon} - \frac{\pi}{4}\right) + O(\epsilon^\alpha) \right]$$
$$y_{in}(x) = \frac{B \sqrt{\epsilon}}{x^{1/4}} \left[ \cos \left(\frac{\sqrt{x}}{\epsilon} - \frac{\pi}{4}\right) + O(\epsilon^\alpha) \right]$$
The matching of these solutions should give $B = \frac{1}{\sqrt{\epsilon}} \left( 1 + O(\epsilon^\alpha) \right)$ and $A = O(\epsilon^\alpha) $. How to prove this?
If we introduce the new matching variable $x = \epsilon^\beta \tilde{x}$, then it will be in the overlap region provided $2(1-\alpha) > \beta > \frac{2}{3} (1+ \alpha) $. The matching condition should then say that the difference $y_{out} - y_{in}$ expressed in the new variable $\tilde{x}$ tends to $0$ as $\epsilon \to 0$ (and $\tilde{x}$ is fixed) for all $\beta$ as above, right? This difference reads
$$\frac{\cos{\theta_\epsilon}}{\tilde{x}^{1/4}} \cdot \frac{1 - B\sqrt{\epsilon}}{\epsilon^{\beta/4}} + \frac{A \sin{\theta_\epsilon}}{\tilde{x}^{1/4}\epsilon^{\beta/4}} + B \cdot O(\epsilon^{\alpha + \frac{1}{2} - \frac{\beta}{4}}) + O(\epsilon^{\alpha - \frac{\beta}{4}})$$
where $\theta_\epsilon = \sqrt{\tilde{x}} \epsilon^{\frac{\beta}{2} - 1} - \frac{\pi}{4} \to \infty$ as $\epsilon \to 0$. The last term goes to $0$, because $\alpha - \frac{\beta}{4} > 0$. The remaining terms will also go to $0$ if
$$1 - B\sqrt{\epsilon} = o(\epsilon^{\beta/4}), \quad A = o (\epsilon^{\beta/4})$$
How to conclude from here?