`lim_(xrarr1)(1-x)tan((pix)/2)` is equal to

Options

A. `(pi)/2`
B. `2/(pi)`
C. `-(pi)/2`
D. `-2/(pi)`

Explanation

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

### Step 1: Substitute the limit
First, we substitute \( x = 1 \) directly into the expression:
\[ \lim_{x \to 1} (1 - x) \tan\left(\frac{\pi x}{2}\right) = (1 - 1) \tan\left(\frac{\pi \cdot 1}{2}\right) = 0 \cdot \tan\left(\frac{\pi}{2}\right) \]
Since \( \tan\left(\frac{\pi}{2}\right) \) is undefined (approaches infinity), we have an indeterminate form \( 0 \cdot \infty \).

### Step 2: Rewrite the expression
To resolve the indeterminate form, we can rewrite the expression:
\[ \lim_{x \to 1} (1 - x) \tan\left(\frac{\pi x}{2}\right) = \lim_{x \to 1} \frac{(1 - x)}{\cot\left(\frac{\pi x}{2}\right)} \]
This transforms our limit into a \( \frac{0}{0} \) form as \( x \to 1 \).

### Step 3: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:
\[ \lim_{x \to 1} \frac{(1 - x)}{\cot\left(\frac{\pi x}{2}\right)} = \lim_{x \to 1} \frac{-1}{-\csc^2\left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2}} = \lim_{x \to 1} \frac{1}{\csc^2\left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2}} \]

### Step 4: Evaluate the limit
Now we need to evaluate \( \csc^2\left(\frac{\pi x}{2}\right) \) as \( x \to 1 \):
\[ \csc\left(\frac{\pi x}{2}\right) = \frac{1}{\sin\left(\frac{\pi x}{2}\right)} \quad \text{and as } x \to 1, \sin\left(\frac{\pi x}{2}\right) \to 1 \]
Thus, \( \csc^2\left(\frac{\pi x}{2}\right) \to 1 \).

Substituting this back into our limit gives:
\[ \lim_{x \to 1} \frac{1}{\csc^2\left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2}} = \frac{1}{1 \cdot \frac{\pi}{2}} = \frac{2}{\pi} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 1} (1 - x) \tan\left(\frac{\pi x}{2}\right) = \frac{2}{\pi} \]