Evaluating a 0/0 limit with L'Hopital's Rule

Substituting $x = \pi/2$ gives numerator $1 - \sin(\pi/2) = 0$ and denominator $\cos^2(\pi/2) = 0$, confirming the $0/0$ form. Differentiating separately, the numerator's derivative is $-\cos(x)$, and the denominator's derivative (using the chain rule on $\cos^2(x)$) is $-2\cos(x)\sin(x)$. This gives the new limit $\lim_{x \to \pi/2} \dfrac{-\cos(x)}{-2\cos(x)\sin(x)}$.

Since $\cos(x)$ appears in both numerator and denominator (and is nonzero near $\pi/2$, except exactly at $\pi/2$), it cancels algebraically, leaving $\lim_{x \to \pi/2} \dfrac{1}{2\sin(x)}$. Substituting $x = \pi/2$ gives $\dfrac{1}{2(1)} = \dfrac{1}{2}$, so the limit equals $\frac{1}{2}$.