If `f(x)=[tan((pi)/(4)+x)]^((1)/(x)),xne0" " f(x)=k , "x=0`, `" is continuous at x=0 "` Then k= . . . .

Options

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

Explanation

To find the value of \( k \) such that the function \( f(x) = \left[ \tan\left(\frac{\pi}{4} + x\right) \right]^{\frac{1}{x}} \) is continuous at \( x = 0 \), we need to ensure that:

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

### Step 1: Calculate \( f(x) \) for \( x \neq 0 \)

Given:

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

### Step 2: Simplify \( \tan\left(\frac{\pi}{4} + x\right) \)

Using the tangent addition formula:

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

### Step 3: Substitute into \( f(x) \)

Thus,

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

### Step 4: Rewrite \( f(x) \)

We can express \( f(x) \) as:

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

### Step 5: Take the limit as \( x \to 0 \)

We need to evaluate:

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

### Step 6: Use the exponential limit

We can use the fact that:

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

### Step 7: Analyze the components

As \( x \to 0 \):

- \( \tan(x) \approx x \)
- Therefore, \( \tan(x) \to 0 \)

### Step 8: Substitute into the limit

Now we can rewrite our limit:

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

### Step 9: Evaluate each part

1. For \( \left(1 + \tan(x)\right)^{\frac{1}{x}} \):

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

2. For \( \left(1 - \tan(x)\right)^{-\frac{1}{x}} \):

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

### Step 10: Combine the results

Combining these results gives:

\[ \lim_{x \to 0} f(x) = e \cdot \frac{1}{e} = e^2 \]

### Conclusion

Thus, for \( f(x) \) to be continuous at \( x = 0 \), we have:

\[ k = e^2 \]