Please help me for the limitation below, I tried using L'Hospital's Rule but it seems not to have helped. The square root term at the denominator is repeated after taking derivatives.
\begin{align} \lim _{\phi\rightarrow\phi_0+\pi} \frac{2j\sin{[ka\sin{(\phi-\phi_0)}}]}{2\sqrt{2}jka\sqrt{1-\sin{\frac{\phi-\phi_0}{2}}}} \end{align} Using L'Hospital rule, one gets \begin{align} \lim _{\phi\rightarrow\phi_0+\pi} \frac{2j\cos{[ka\sin{(\phi-\phi_0)}}]ka\cos{(\phi-\phi_0)}}{2\sqrt{2}jka\frac{-\frac{1}{2}\cos{\frac{\phi-\phi_0}{2}}}{2\sqrt{1-\sin{\frac{\phi-\phi_0}{2}}}}} \end{align} The norminator is converged to $-2jka$, however the denorminator is in the form of $\frac{0}{0}$, then I continued determining the convergence of the denorminator by using L'Hospital as following
\begin{align} \lim _{\phi\rightarrow\phi_0+\pi}2\sqrt{2}jka\frac{-\frac{1}{2}\cos{\frac{\phi-\phi_0}{2}}}{2\sqrt{1-\sin{\frac{\phi-\phi_0}{2}}}} = \lim _{\phi\rightarrow\phi_0+\pi}2\sqrt{2}jka \frac{\frac{1}{4}\sin{\frac{\phi-\phi_0}{2}}}{2\frac{-\frac{1}{2}\cos{\frac{\phi-\phi_0}{2}}}{2\sqrt{1-\sin{\frac{\phi-\phi_0}{2}}}}} \end{align} Again, you can see that the term, $\frac{\cos{\frac{\phi-\phi_0}{2}}}{\sqrt{1-\sin{\frac{\phi-\phi_0}{2}}}}$, is repeated.
Thank you so much!
Use equivalent infinitesimal. Write $\phi-\phi_0$ as $x+\pi$, then $\phi\to\phi_0+\pi\iff x\to0$. Consider what is the equivalent infinitesimal of $\sin[ka\sin(x+\pi)]$? And note that $\sin^2\frac{x}{4}=\frac{1-\cos\frac{x}{2}}{2}$.