`f(x) = {{:(("tan"((pi)/(4)+x))^(1//x)",",x ne 0),(k",",x = 0):}` for what value of k, f(x) is continuous at x = 0 ?

Answer

To determine the value of \( k \) for which the function

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

is continuous at \( x = 0 \), we need to ensure that the left-hand limit, right-hand limit, and the value of the function at \( x = 0 \) are all equal.

### Step 1: Find the Left-Hand Limit as \( x \to 0^- \)

We need to calculate:

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

### Step 2: Find the Right-Hand Limit as \( x \to 0^+ \)

Similarly, we calculate:

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

### Step 3: Calculate the Limit

Since both limits are the same, we can calculate just one of them. We know that:

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

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

\[ \tan\left(\frac{\pi}{4} + x\right) \to \frac{1 + x}{1 - x} \to 1 \text{ as } x \to 0 \]

So, we can rewrite our limit:

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

This is an indeterminate form.

### Step 4: Use the Exponential Limit Trick

We can use the fact that \( 1^{\infty} \) can be evaluated using the exponential limit:

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

### Step 5: Find the Derivative

Now we need to calculate:

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

Using L'Hôpital's Rule, we differentiate the numerator and denominator:

1. The derivative of the numerator \( \tan\left(\frac{\pi}{4} + x\right) - 1 \) is \( \sec^2\left(\frac{\pi}{4} + x\right) \).
2. The derivative of the denominator \( x \) is \( 1 \).

Thus, we have:

\[ \lim_{x \to 0} \sec^2\left(\frac{\pi}{4} + x\right) = \sec^2\left(\frac{\pi}{4}\right) = 2 \]

### Step 6: Final Calculation

So, we have:

\[ \lim_{x \to 0} \tan\left(\frac{\pi}{4} + x\right)^{\frac{1}{x}} = e^2 \]

### Step 7: Set the Limits Equal

For \( f(x) \) to be continuous at \( x = 0 \):

\[ k = e^2 \]

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

\[ \boxed{e^2} \]

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