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We can change variable from x to $\sin(x)$

then

$\displaystyle \lim_{u \rightarrow 1} \dfrac{u-u^{u}}{1-u+\ln(u)}$

Then, we apply L'Hospital's rule, 2 times

$\displaystyle \lim _{ u\rightarrow 1 }{ \frac { -{ u }^{ u }\left( 1+\ln { u } \right) -{ u }^{ u }\ln { u } \left( 1+\ln { u } \right) -\frac { { u }^{ u } }{ u } }{ -\frac { 1 }{ { u }^{ 2 } } } } =1+1=2$

Simple,

$\displaystyle \begin{aligned} \lim_{x\to\frac\pi2} \frac{\sin(x) - (\sin(x))^{\sin(x)}}{1 - \sin(x) + \ln(\sin(x))} &= \lim_{x \to \frac{\pi}{2}} \frac{\frac{\mathrm{d}^2}{\mathrm{d}x^2} ~~\left[\sin(x) - (\sin(x))^{\sin(x)}\right]}{ \frac{\mathrm{d}^2}{\mathrm{d}x^2} ~~\left[1 - \sin(x) + \ln\sin(x)\right]}\\ &=\lim_{x \to \frac{\pi}{2}}\frac{ -\left[(\sin x) ^ {\sin x} (1 + \ln \sin x)^2 + \frac{ (\sin x)^{\sin x}}{\sin x} \right]}{\frac{-1}{\sin^2 x}}\\ &= \frac{-1-1}{-1} \\& = \boxed{2}\end{aligned}$