Existence is key

$\begin{aligned} L & = \lim_{x \to \frac \pi 2} \frac {\sin \left(x-\frac \pi 2\right)}{x^2 - \pi x + \frac {\pi^2}4} = \lim_{x \to \frac \pi 2} \frac {\sin \left(x-\frac \pi 2\right)}{\left(x-\frac \pi 2\right)^2} & \small \blue{\text{Let }u = x - \frac \pi 2} \\ & = \lim_{u \to 0} \frac {\sin u}u \cdot \frac 1u = \lim_{u \to 0} \frac 1u \end{aligned}$

$\displaystyle \implies L = \lim_{u \to 0} \dfrac 1u \implies \begin{cases} \displaystyle L^- = \lim_{u \to 0^-} \dfrac 1u = - \infty \\ \displaystyle L^+ = \lim_{u \to 0^+} \dfrac 1u = \infty \end{cases}$

Since $L^- \ne L^+$ , $L$ is undefined .

I'm not going to explain but the denominator approaches $0$ and division by $0$ is impossible so the answer is $-0.12345$

When evaluating this limit from $\left(\cfrac{\pi}{2}\right)^{+}$ , it comes out as $\infty$ . However, evaluating it from $\left(\cfrac{\pi}{2}\right)^{-}$ returns $-\infty$ . Since these are not the same, the limit $\green{\boxed{\text{doesn't exist}}}$

$\lim_{x \rightarrow \frac{\pi}{2}}\frac{\sin(x-\frac{\pi}{2})}{x^2-\pi x+\frac{\pi^2}{4}} =\lim_{x \rightarrow \frac{\pi}{2}} \frac{\sin(x)}{2x-\pi}= \frac{1}{0}$

Which is undefined