`Lim_(x to pi/2) (1+cos 2x)/(pi-2x)^(2)`

Answer

To solve the limit \( \lim_{x \to \frac{\pi}{2}} \frac{1 + \cos(2x)}{(\pi - 2x)^2} \), we will follow these steps:

### Step 1: Substitute \( x = \frac{\pi}{2} + h \)

As \( x \) approaches \( \frac{\pi}{2} \), we can let \( h \) approach \( 0 \) by substituting \( x = \frac{\pi}{2} + h \). This gives us:

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

### Step 2: Simplify the expression

Now we simplify the expression:

\[ \cos(2(\frac{\pi}{2} + h)) = \cos(\pi + 2h) = -\cos(2h) \]

Thus, the limit becomes:

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

### Step 3: Simplify the denominator

The denominator simplifies to:

\[ (-2h)^2 = 4h^2 \]

So now we have:

\[ \lim_{h \to 0} \frac{1 - \cos(2h)}{4h^2} \]

### Step 4: Use the trigonometric identity

We know from trigonometric identities that:

\[ 1 - \cos(2h) = 2\sin^2(h) \]

Substituting this into our limit gives:

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

### Step 5: Apply the limit property

Using the limit property \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \), we can rewrite the limit:

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

Thus, the limit becomes:

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

### Final Answer

Therefore, the limit is:

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

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