Evaluate : `lim_(x to(pi)/(2))(tan 2 x)/( x - (pi)/(2))`

Answer

To evaluate the limit

\[ \lim_{x \to \frac{\pi}{2}} \frac{\tan(2x)}{x - \frac{\pi}{2}}, \]

we will follow these steps:

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

\[ \tan(2x) = \tan(2 \cdot \frac{\pi}{2}) = \tan(\pi) = 0, \]
and

\[ x - \frac{\pi}{2} = \frac{\pi}{2} - \frac{\pi}{2} = 0. \]

Thus, we have the form \( \frac{0}{0} \).

**Hint:** When substituting the limit value results in \( \frac{0}{0} \), it indicates that we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Since we have the indeterminate form \( \frac{0}{0} \), 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.

Here, let \( f(x) = \tan(2x) \) and \( g(x) = x - \frac{\pi}{2} \).

### Step 3: Differentiate the numerator and denominator
Now, we differentiate \( f(x) \) and \( g(x) \):

1. **Differentiate \( f(x) = \tan(2x) \)**:
\[ f'(x) = \sec^2(2x) \cdot 2 = 2 \sec^2(2x) \]

2. **Differentiate \( g(x) = x - \frac{\pi}{2} \)**:
\[ g'(x) = 1 \]

### Step 4: Substitute into the limit
Now we substitute these derivatives back into the limit:

\[ \lim_{x \to \frac{\pi}{2}} \frac{f'(x)}{g'(x)} = \lim_{x \to \frac{\pi}{2}} \frac{2 \sec^2(2x)}{1} = 2 \sec^2(2 \cdot \frac{\pi}{2}) = 2 \sec^2(\pi). \]

### Step 5: Evaluate \( \sec^2(\pi) \)
We know that:

\[ \sec(\pi) = \frac{1}{\cos(\pi)} = \frac{1}{-1} = -1. \]

Thus,

\[ \sec^2(\pi) = (-1)^2 = 1. \]

### Step 6: Final result
Now, substituting this back into our limit gives:

\[ \lim_{x \to \frac{\pi}{2}} \frac{\tan(2x)}{x - \frac{\pi}{2}} = 2 \cdot 1 = 2. \]

Therefore, the final answer is:

\[ \boxed{2}. \]

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