The given limit problem is:

$\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2}$

Step 1: Substituting $x = \frac{\pi}{2}$

First, check the form of the limit by directly substituting $x = \frac{\pi}{2}$:

• $\sin\left( \frac{\pi}{2} \right) = 1$,
• $\frac{\pi}{2} - \frac{\pi}{2} = 0$.

Substituting into the expression gives:

$\frac{1 - \sin\left( \frac{\pi}{2} \right)}{\left( \frac{\pi}{2} - \frac{\pi}{2} \right)^2} = \frac{1 - 1}{0^2} = \frac{0}{0}$

This is an indeterminate form $\frac{0}{0}$, so we can apply L'Hôpital's Rule.

Step 2: Applying L'Hôpital's Rule

L'Hôpital's Rule states that if a limit results in an indeterminate form $\frac{0}{0}$, we can differentiate the numerator and the denominator with respect to $x$, and then re-evaluate the limit.

The expression is:

$\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2}$

Differentiate the numerator:

The derivative of $1 - \sin x$ with respect to $x$ is:

$-\cos x$

Differentiate the denominator:

The derivative of $\left( \frac{\pi}{2} - x \right)^2$ with respect to $x$ is:

$2\left( \frac{\pi}{2} - x \right) \cdot (-1) = -2 \left( \frac{\pi}{2} - x \right)$

So, applying L'Hôpital's Rule gives:

$\lim_{x \to \frac{\pi}{2}} \frac{-\cos x}{-2 \left( \frac{\pi}{2} - x \right)}$

Simplifying this:

$\lim_{x \to \frac{\pi}{2}} \frac{\cos x}{2 \left( \frac{\pi}{2} - x \right)}$

Step 3: Evaluating the Limit

Now substitute $x = \frac{\pi}{2}$ into the expression. Since $\cos \left( \frac{\pi}{2} \right) = 0$, we get:

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

Again, this results in an indeterminate form, so we apply L'Hôpital's Rule a second time.

Step 4: Second Application of L'Hôpital's Rule

Differentiate the numerator and denominator again.

Numerator:

The derivative of $\cos x$ is:

$-\sin x$

Denominator:

The derivative of $2 \left( \frac{\pi}{2} - x \right)$ is:

$-2$

So the new limit is:

$\lim_{x \to \frac{\pi}{2}} \frac{-\sin x}{-2} = \lim_{x \to \frac{\pi}{2}} \frac{\sin x}{2}$

Now substitute $x = \frac{\pi}{2}$ into this expression:

$\frac{\sin \left( \frac{\pi}{2} \right)}{2} = \frac{1}{2}$

Final Answer:

$\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2} = \frac{1}{2}$

Let me know if you have any questions or need more details.

Here are 5 related questions to expand your understanding:

1. How does L'Hôpital's Rule help in solving indeterminate forms?
2. What other techniques can be used to solve limits besides L'Hôpital's Rule?
3. Can the limit be evaluated using series expansions around $x = \frac{\pi}{2}$?
4. What are the conditions under which L'Hôpital's Rule can be applied?
5. What would happen if we tried to solve this limit using a Taylor series expansion?

Tip: Always check if a limit is indeterminate before applying L'Hôpital’s Rule—other methods like factorization or series expansion might be simpler!