We need to evaluate the limit:

$ L = \lim_{x \to \pi/2} \frac{\sin(\cos (x)) - \cos (x)}{\frac{\pi}{2} - x} $

Limit Indeterminate Form Identification

As $x \to \pi/2$:

• $\cos(x) \to \cos(\pi/2) = 0$
• Numerator $\to \sin(0) - 0 = 0$
• Denominator $\to \pi/2 - \pi/2 = 0$

The limit is in the indeterminate form $\frac{0}{0}$.

Applying Substitution

Let $y = \frac{\pi}{2} - x$. As $x \to \pi/2$, we have $y \to 0$. From this substitution, $x = \frac{\pi}{2} - y$. We also know that $\cos(x) = \cos(\frac{\pi}{2} - y) = \sin(y)$.

Substituting these into the limit expression:

$ L = \lim_{y \to 0} \frac{\sin(\sin y) - \sin y}{y} $

Evaluating the Transformed Limit

This is still in the $\frac{0}{0}$ form as $y \to 0$. We can rewrite the expression:

$ L = \lim_{y \to 0} \left( \frac{\sin(\sin y)}{y} - \frac{\sin y}{y} \right) $

We can manipulate the first term:

$ \frac{\sin(\sin y)}{y} = \frac{\sin(\sin y)}{\sin y} \cdot \frac{\sin y}{y} $

Now, we use the standard limit property $\lim_{z \to 0} \frac{\sin z}{z} = 1$.

• Let $z = \sin y$. As $y \to 0$, $z \to 0$. Therefore, $\lim_{y \to 0} \frac{\sin(\sin y)}{\sin y} = \lim_{z \to 0} \frac{\sin z}{z} = 1$.
• Also, $\lim_{y \to 0} \frac{\sin y}{y} = 1$.

Substitute these back into the expression for L:

$ L = \lim_{y \to 0} \left( \frac{\sin(\sin y)}{\sin y} \cdot \frac{\sin y}{y} - \frac{\sin y}{y} \right) $

$ L = (1 \cdot 1) - 1 $

$ L = 1 - 1 = 0 $

Final Solution

The value of the limit is 0.