What is the limit?

This is a nice example of a $1^{\infty}$ case of L'Hôpital's rule. As usual, the easiest approach is to convert the limit into a quotient: we have $\begin{aligned} \log \left((2-\sin x+\cos x)^{\tan x} \right) &= \tan x \cdot \log (2-\sin x+\cos x) \\ &= \frac{\log (2-\sin x+\cos x)}{\cot x} \end{aligned}$

Both the numerator and denominator tend to zero in the limit, so we can now apply L'Hôpital's rule to the logarithm:

$\begin{aligned} \lim_{x \to \frac{\pi}{2}} \log \left((2-\sin x+\cos x)^{\tan x} \right) &= \lim_{x \to \frac{\pi}{2}} \frac{(\cos x + \sin x) \sin^2 x}{2-\sin x + \cos x} \\ &= 1\end{aligned}$

so that the required limit is $\boxed{e}$ .