The value of `lim_(x rarr a) (2- a/x)^(tan ((pi x)/(2a))) ` is:

Options

A. `e^(-1//pi)`
B. `e^(2//pi)`
C. `e^(-2//pi)`
D. `e^(1//pi)`

Explanation

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

### Step 1: Rewrite the expression
We start by rewriting the expression inside the limit:
\[ L = \lim_{x \to a} \left(2 - \frac{a}{x}\right)^{\tan\left(\frac{\pi x}{2a}\right)} \]
We can express \(2\) as \(1 + 1\):
\[ L = \lim_{x \to a} \left(1 + \left(1 - \frac{a}{x}\right)\right)^{\tan\left(\frac{\pi x}{2a}\right)} \]

### Step 2: Check the form of the limit
Now, we evaluate the limit as \(x\) approaches \(a\):
\[ \frac{a}{x} \to 1 \quad \text{as } x \to a \Rightarrow 1 - \frac{a}{x} \to 0 \]
Thus, we have:
\[ L = \lim_{x \to a} \left(1 + \left(1 - \frac{a}{x}\right)\right)^{\tan\left(\frac{\pi x}{2a}\right)} \to 1^{\infty} \]
This is an indeterminate form \(1^{\infty}\).

### Step 3: Apply the exponential limit theorem
Using the standard limit form for \(1^{\infty}\):
\[ L = e^{\lim_{x \to a} \left(1 - \frac{a}{x}\right) \tan\left(\frac{\pi x}{2a}\right)} \]

### Step 4: Substitute \(x = a + h\)
To evaluate the limit, we can substitute \(x = a + h\) where \(h \to 0\):
\[ 1 - \frac{a}{x} = 1 - \frac{a}{a + h} = \frac{h}{a + h} \]
And,
\[ \tan\left(\frac{\pi x}{2a}\right) = \tan\left(\frac{\pi (a + h)}{2a}\right) = \tan\left(\frac{\pi}{2} + \frac{\pi h}{2a}\right) = -\cot\left(\frac{\pi h}{2a}\right) \]

### Step 5: Substitute back into the limit
Now substituting back into the limit:
\[ L = e^{\lim_{h \to 0} \frac{h}{a + h} \cdot \left(-\cot\left(\frac{\pi h}{2a}\right)\right)} \]

### Step 6: Evaluate the limit
Using the fact that \(\cot\left(x\right) = \frac{\cos(x)}{\sin(x)}\) and \(\sin(x) \sim x\) as \(x \to 0\):
\[ \cot\left(\frac{\pi h}{2a}\right) \sim \frac{2a}{\pi h} \quad \text{as } h \to 0 \]
Thus,
\[ L = e^{\lim_{h \to 0} \frac{h}{a + h} \cdot \left(-\frac{2a}{\pi h}\right)} = e^{\lim_{h \to 0} -\frac{2a}{\pi(a + h)}} \]
As \(h \to 0\), this limit simplifies to:
\[ L = e^{-\frac{2a}{\pi a}} = e^{-\frac{2}{\pi}} \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{e^{-\frac{2}{\pi}}} \]