Evaluate the following limits: `Lim_( x to pi/2 )(1-sin x)/((pi/2-x)^(2))`

Answer

To evaluate the limit

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

we will follow these steps:

### Step 1: Identify the form of the limit
First, we substitute \( x = \frac{\pi}{2} \) directly into the limit:

\[ 1 - \sin\left(\frac{\pi}{2}\right) = 1 - 1 = 0, \]
and
\[ \left(\frac{\pi}{2} - \frac{\pi}{2}\right)^2 = 0^2 = 0. \]

This gives us a \( \frac{0}{0} \) indeterminate form, so we need to simplify it.

### Step 2: Change of variable
Let \( x = \frac{\pi}{2} - h \). As \( x \to \frac{\pi}{2} \), \( h \to 0 \). Now, we rewrite the limit in terms of \( h \):

\[ \lim_{h \to 0} \frac{1 - \sin\left(\frac{\pi}{2} - h\right)}{h^2}. \]

### Step 3: Simplify the sine function
Using the identity \( \sin\left(\frac{\pi}{2} - h\right) = \cos h \), we can rewrite the limit:

\[ \lim_{h \to 0} \frac{1 - \cos h}{h^2}. \]

### Step 4: Use trigonometric identities
We can use the identity \( 1 - \cos h = 2 \sin^2\left(\frac{h}{2}\right) \):

\[ \lim_{h \to 0} \frac{2 \sin^2\left(\frac{h}{2}\right)}{h^2}. \]

### Step 5: Rewrite the limit
Next, we rewrite the limit:

\[ \lim_{h \to 0} 2 \cdot \frac{\sin^2\left(\frac{h}{2}\right)}{h^2} = \lim_{h \to 0} 2 \cdot \left(\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right)^2 \cdot \frac{1}{\left(\frac{1}{2}\right)^2}. \]

### Step 6: Apply the limit
We know that

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1. \]

Thus,

\[ \lim_{h \to 0} \left(\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right)^2 = 1. \]

### Step 7: Calculate the final value
Putting it all together:

\[ \lim_{h \to 0} 2 \cdot 1 \cdot \frac{1}{\left(\frac{1}{2}\right)^2} = 2 \cdot 1 \cdot 4 = 8. \]

Thus, the limit evaluates to:

\[ \frac{1}{2}. \]

### Final Answer
Therefore,

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

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