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We are given the following limit problem, which involves trigonometric functions:

\[
\lim_{{x \to \frac{\pi}{2}}} \frac{1 - \sin(x)}{x - \frac{\pi}{2}}
\]

To solve this, we will use a change of variable to simplify the expression.

### Step 1: Change of Variable

Let \( u = x - \frac{\pi}{2} \). This means that as \( x \to \frac{\pi}{2} \), we have \( u \to 0 \). So the limit becomes:

\[
\lim_{u \to 0} \frac{1 - \sin\left( u + \frac{\pi}{2} \right)}{u}
\]

### Step 2: Simplify the Sine Expression

Using the identity for \( \sin\left( u + \frac{\pi}{2} \right) \), we know that:

\[
\sin\left( u + \frac{\pi}{2} \right) = \cos(u)
\]

Thus, the limit simplifies to:

\[
\lim_{u \to 0} \frac{1 - \cos(u)}{u}
\]

### Step 3: Apply Known Limit

We can now use the known trigonometric limit:

\[
\lim_{u \to 0} \frac{1 - \cos(u)}{u} = 0
\]

Therefore, the original limit is:

\[
\boxed{0}
\]

Would you like more details on any step, or do you have any questions?

Here are 5 related questions to expand your understanding:
1. How can you apply L'Hopital's Rule to evaluate this limit?
2. What other trigonometric identities can help simplify similar limits?
3. How would the limit change if the function was \( \cos(x) \) instead of \( \sin(x) \)?
4. How does the change of variable method compare to direct substitution in limits?
5. What is the significance of the limit \( \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} \)?

**Tip**: Whenever dealing with trigonometric limits near \( \frac{\pi}{2} \) or \( 0 \), using small-angle approximations or trigonometric identities can greatly simplify the problem.

Evaluar, haciendo cambio de variable, el límite trigonométrico: lim (x -> pi/2) (1 - sin(x)) / (x - pi/2)

Learn how to evaluate the trigonometric limit lim (x -> pi/2) (1 - sin(x)) / (x - pi/2) using a change of variable. This solution covers essential calculus concepts such as limit theorems and trigonometric identities. Ideal for undergraduate students.

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