To solve the limit

$\lim_{x \to 0} \frac{\sin\left(\frac{\pi}{2} + x\right) - \sin(x)}{x}$

we'll start by using the trigonometric identity for the sine of a sum:

$\sin\left(\frac{\pi}{2} + x\right) = \sin\left(\frac{\pi}{2}\right)\cos(x) + \cos\left(\frac{\pi}{2}\right)\sin(x)$

Since $\sin\left(\frac{\pi}{2}\right) = 1$ and $\cos\left(\frac{\pi}{2}\right) = 0$, the expression simplifies to:

$\sin\left(\frac{\pi}{2} + x\right) = \cos(x)$

Thus, the original limit becomes:

$\lim_{x \to 0} \frac{\cos(x) - \sin(x)}{x}$

To resolve the indeterminate form $\frac{0}{0}$ as $x \to 0$, we use L'Hôpital's Rule. According to the rule:

$\lim_{x \to 0} \frac{\cos(x) - \sin(x)}{x} = \lim_{x \to 0} \frac{-\sin(x) - \cos(x)}{1}$

Evaluating at $x = 0$:

$-\sin(0) - \cos(0) = -0 - 1 = -1$

Thus, the limit is:

$\boxed{-1}$

Would you like to explore more about this problem or any related topics?

Here are some related questions to think about:

1. What are other applications of L'Hôpital's Rule?
2. How does the Taylor series approximation help in evaluating limits?
3. What is the geometric interpretation of