Evaluate the following limit :
`Lim_(x to pi/2) (1+cos 2x)/(pi-2x)^(2)`

Answer

To evaluate the limit

\[ \lim_{x \to \frac{\pi}{2}} \frac{1 + \cos(2x)}{(\pi - 2x)^2}, \]

we will follow these steps:

### Step 1: Direct Substitution
First, we try to substitute \( x = \frac{\pi}{2} \) directly into the limit.

\[ 1 + \cos(2 \cdot \frac{\pi}{2}) = 1 + \cos(\pi) = 1 - 1 = 0, \]
and
\[ (\pi - 2 \cdot \frac{\pi}{2})^2 = (\pi - \pi)^2 = 0^2 = 0. \]

Since both the numerator and denominator approach 0, we have an indeterminate form \( \frac{0}{0} \).

### Step 2: Rewrite the Limit
To resolve the indeterminate form, we can rewrite \( x \) as \( x = \frac{\pi}{2} - h \) where \( h \to 0 \) as \( x \to \frac{\pi}{2} \). Thus, we rewrite the limit:

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

### Step 3: Simplify the Expression
Now, simplify the expressions:

1. The numerator becomes:
\[ 1 + \cos(\pi - 2h) = 1 - \cos(2h) = 1 - (1 - 2\sin^2(h)) = 2\sin^2(h). \]

2. The denominator simplifies to:
\[ \pi - \pi + 2h = 2h \quad \text{so} \quad (2h)^2 = 4h^2. \]

### Step 4: Substitute Back into the Limit
Now we substitute back into the limit:

\[ \lim_{h \to 0} \frac{2\sin^2(h)}{4h^2} = \lim_{h \to 0} \frac{\sin^2(h)}{2h^2}. \]

### Step 5: Use the Limit Property
Using the known limit \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \), we can rewrite:

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

Thus, we have:

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

### Final Answer
Therefore, the limit evaluates to:

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