`lim_(x to pi/2) (tan2x)/(x-(pi)/(2))`

Explanation

To solve the limit \( \lim_{x \to \frac{\pi}{2}} \frac{\tan(2x)}{x - \frac{\pi}{2}} \), we can follow these steps:

### Step 1: Substitute \( x = \frac{\pi}{2} \)
First, we substitute \( x = \frac{\pi}{2} \) into the limit expression:

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

This gives us the indeterminate form \( \frac{0}{0} \).

**Hint:** When you get a \( \frac{0}{0} \) form, it indicates that you can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Since we have the 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 exists.

### Step 3: Differentiate the numerator and denominator
Now we differentiate the numerator and denominator separately:

- The numerator \( f(x) = \tan(2x) \):
\[ f'(x) = \sec^2(2x) \cdot \frac{d}{dx}(2x) = \sec^2(2x) \cdot 2 = 2 \sec^2(2x) \]

- The denominator \( g(x) = x - \frac{\pi}{2} \):
\[ g'(x) = 1 \]

### Step 4: Rewrite the limit
Now we can rewrite the limit using the derivatives:

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

### Step 5: Substitute \( x = \frac{\pi}{2} \) again
Now we substitute \( x = \frac{\pi}{2} \) into the new limit expression:

\[ = 2 \sec^2(2 \cdot \frac{\pi}{2}) = 2 \sec^2(\pi) \]

Since \( \sec(\pi) = \frac{1}{\cos(\pi)} = -1 \), we have:

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

Thus, the limit evaluates to:

\[ = 2 \cdot 1 = 2 \]

### Final Answer
\[ \lim_{x \to \frac{\pi}{2}} \frac{\tan(2x)}{x - \frac{\pi}{2}} = 2 \]

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