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

Answer

To evaluate the limit

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

we will follow these steps:

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

First, we substitute \( x = \frac{\pi}{2} \) into the expression to check if it results in an indeterminate form.

\[ \cos\left(\frac{\pi}{2}\right) = 0 \implies 2^{-\cos\left(\frac{\pi}{2}\right)} = 2^0 = 1 \]

Thus, the numerator becomes:

\[ 2^{-\cos\left(\frac{\pi}{2}\right)} - 1 = 1 - 1 = 0 \]

The denominator also becomes:

\[ \frac{\pi}{2}\left(\frac{\pi}{2} - \frac{\pi}{2}\right) = \frac{\pi}{2} \cdot 0 = 0 \]

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

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

if the limit on the right side exists.

Let \( f(x) = 2^{-\cos x} - 1 \) and \( g(x) = x(x - \frac{\pi}{2}) \).

### Step 3: Differentiate the Numerator and Denominator

**Numerator:**

To differentiate \( f(x) = 2^{-\cos x} - 1 \), we use the chain rule:

\[ f'(x) = \frac{d}{dx}(2^{-\cos x}) = 2^{-\cos x} \cdot \ln(2) \cdot \sin x \]

**Denominator:**

For the denominator \( g(x) = x(x - \frac{\pi}{2}) \):

\[ g'(x) = \frac{d}{dx}(x^2 - \frac{\pi}{2}x) = 2x - \frac{\pi}{2} \]

### Step 4: Rewrite the Limit Using Derivatives

Now we rewrite the limit using the derivatives:

\[ \lim_{x \to \frac{\pi}{2}} \frac{f'(x)}{g'(x)} = \lim_{x \to \frac{\pi}{2}} \frac{2^{-\cos x} \cdot \ln(2) \cdot \sin x}{2x - \frac{\pi}{2}} \]

### Step 5: Substitute \( x = \frac{\pi}{2} \) Again

Now we substitute \( x = \frac{\pi}{2} \):

**Numerator:**

\[ f'\left(\frac{\pi}{2}\right) = 2^{-\cos\left(\frac{\pi}{2}\right)} \cdot \ln(2) \cdot \sin\left(\frac{\pi}{2}\right) = 2^0 \cdot \ln(2) \cdot 1 = \ln(2) \]

**Denominator:**

\[ g'\left(\frac{\pi}{2}\right) = 2\left(\frac{\pi}{2}\right) - \frac{\pi}{2} = \pi - \frac{\pi}{2} = \frac{\pi}{2} \]

### Step 6: Calculate the Limit

Now we can compute the limit:

\[ \lim_{x \to \frac{\pi}{2}} \frac{f'(x)}{g'(x)} = \frac{\ln(2)}{\frac{\pi}{2}} = \frac{2\ln(2)}{\pi} \]

Thus, the final answer is:

\[ \boxed{\frac{2\ln(2)}{\pi}} \]