`lim_(x rarr 0)tan(pi/4+x)^(1/x)=`

Options

A. e
B. `e^2`
C. `e^4`
D. `1/e`

Answer

To solve the limit \( \lim_{x \to 0} \tan\left(\frac{\pi}{4} + x\right)^{\frac{1}{x}} \), we can follow these steps:

### Step 1: Rewrite the limit
We start by rewriting the expression inside the limit:
\[ \tan\left(\frac{\pi}{4} + x\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan(x)}{1 - \tan\left(\frac{\pi}{4}\right)\tan(x)} \]
Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), we have:
\[ \tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan(x)}{1 - \tan(x)} \]

### Step 2: Substitute into the limit
Now, substituting this back into the limit, we have:
\[ \lim_{x \to 0} \left(\frac{1 + \tan(x)}{1 - \tan(x)}\right)^{\frac{1}{x}} \]

### Step 3: Analyze the limit form
As \( x \to 0 \), \( \tan(x) \to 0 \). Thus, we can express this limit in the form of \( 1 + f(x) \):
\[ \lim_{x \to 0} \left(1 + \frac{2\tan(x)}{1 - \tan(x)}\right)^{\frac{1}{x}} \]
This can be simplified as:
\[ \lim_{x \to 0} \left(1 + 2\tan(x)\right)^{\frac{1}{x}} \]

### Step 4: Use the exponential limit
We know that if \( f(x) \to 0 \) and \( g(x) \to \infty \), then:
\[ \lim_{x \to a} (1 + f(x))^{g(x)} = e^{\lim_{x \to a} f(x)g(x)} \]
Here, we have \( f(x) = 2\tan(x) \) and \( g(x) = \frac{1}{x} \).

### Step 5: Calculate the limit of \( f(x)g(x) \)
Now we compute:
\[ \lim_{x \to 0} 2\tan(x) \cdot \frac{1}{x} \]
Using the fact that \( \tan(x) \sim x \) as \( x \to 0 \):
\[ \lim_{x \to 0} 2 \cdot \frac{\tan(x)}{x} = 2 \cdot 1 = 2 \]

### Step 6: Final limit calculation
Thus, we have:
\[ \lim_{x \to 0} \tan\left(\frac{\pi}{4} + x\right)^{\frac{1}{x}} = e^{2} \]

### Conclusion
The final answer is:
\[ \boxed{e^2} \]