If f(x) = `= {{:([tan ((pi)/(4)+x)]^(1//x),x ne0),(k ,x=0):}`
For what value of k f(x) is continuous at x=0 ?

Options

A. 1
B. 0
C. e
D. `e^(2)`

Answer

To determine the value of \( k \) for which the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that:

\[ \lim_{x \to 0} f(x) = f(0) \]

Given the function:

\[ f(x) = \begin{cases} \tan\left(\frac{\pi}{4} + x\right)^{\frac{1}{x}} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

### Step 1: Calculate \( f(0) \)

From the definition of the function, we have:

\[ f(0) = k \]

### Step 2: Calculate \( \lim_{x \to 0} f(x) \)

We need to find the limit of \( f(x) \) as \( x \) approaches 0. Thus, we evaluate:

\[ \lim_{x \to 0} \tan\left(\frac{\pi}{4} + x\right)^{\frac{1}{x}} \]

### Step 3: Simplify \( \tan\left(\frac{\pi}{4} + x\right) \)

Using the angle addition formula for tangent:

\[ \tan\left(\frac{\pi}{4} + x\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan(x)}{1 - \tan\left(\frac{\pi}{4}\right)\tan(x)} = \frac{1 + \tan(x)}{1 - \tan(x)} \]

As \( x \to 0 \), \( \tan(x) \to x \). Thus:

\[ \tan\left(\frac{\pi}{4} + x\right) \to \frac{1 + x}{1 - x} \]

### Step 4: Calculate the limit

Now we need to find:

\[ \lim_{x \to 0} \left(\frac{1 + x}{1 - x}\right)^{\frac{1}{x}} \]

Using the fact that \( \ln(a^b) = b \ln(a) \):

\[ \ln\left(\left(\frac{1 + x}{1 - x}\right)^{\frac{1}{x}}\right) = \frac{1}{x} \ln\left(\frac{1 + x}{1 - x}\right) \]

### Step 5: Expand using Taylor series

Using the Taylor series expansion for \( \ln(1 + u) \):

\[ \ln(1 + x) \approx x \quad \text{and} \quad \ln(1 - x) \approx -x \]

Thus:

\[ \ln\left(\frac{1 + x}{1 - x}\right) \approx \ln(1 + x) - \ln(1 - x) \approx x - (-x) = 2x \]

### Step 6: Substitute back into the limit

Now substituting back:

\[ \frac{1}{x} \ln\left(\frac{1 + x}{1 - x}\right) \approx \frac{1}{x}(2x) = 2 \]

Thus:

\[ \lim_{x \to 0} \left(\frac{1 + x}{1 - x}\right)^{\frac{1}{x}} = e^2 \]

### Step 7: Set the limit equal to \( k \)

For continuity at \( x = 0 \):

\[ \lim_{x \to 0} f(x) = f(0) \implies e^2 = k \]

### Final Answer

Thus, the value of \( k \) for which \( f(x) \) is continuous at \( x = 0 \) is:

\[ \boxed{e^2} \]