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A function $f(x)$ is continuous at $x=a$ if the limit of the function as $x$ approaches $a$ is equal to the value of the function at $a$. That is, $\lim_{x \to a} f(x) = f(a)$. Here, $a = \frac{\pi}{2}$ and $f(\frac{\pi}{2}) = 1$.
We need to calculate $\lim_{x \to \frac{\pi}{2}} \frac{\cos x}{-x + \frac{\pi}{2}}$. Let $h = x - \frac{\pi}{2}$. As $x \to \frac{\pi}{2}$, $h \to 0$. Then $x = h + \frac{\pi}{2}$.
Substituting the variables, the limit becomes: $$ \lim_{h \to 0} \frac{\cos(h + \frac{\pi}{2})}{- (h + \frac{\pi}{2}) + \frac{\pi}{2}} $$ Using the identity $\cos(\frac{\pi}{2} + h) = -\sin h$, we get: $$ \lim_{h \to 0} \frac{-\sin h}{-h} = \lim_{h \to 0} \frac{\sin h}{h} $$
We know that $\lim_{h \to 0} \frac{\sin h}{h} = 1$. Since the limit equals the function value $f(\frac{\pi}{2}) = 1$, the function is continuous at $x = \frac{\pi}{2}$.
Final Answer: The function is continuous as the limit equals 1.