`lim_(x to 0) {tan (pi/4 +x)}^(1//x)`=

Options

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

Answer

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

### Step 1: Rewrite the limit
We start by rewriting the limit:
\[ L = \lim_{x \to 0} \left( \tan\left(\frac{\pi}{4} + x\right) \right)^{\frac{1}{x}} \]

### Step 2: Use the tangent addition formula
Using the tangent addition formula, we have:
\[ \tan\left(\frac{\pi}{4} + x\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan(x)}{1 - \tan\left(\frac{\pi}{4}\right) \tan(x)} \]
Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), this simplifies to:
\[ \tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan(x)}{1 - \tan(x)} \]

### Step 3: Substitute \(\tan(x)\) with its limit
As \( x \to 0 \), we can use the approximation \( \tan(x) \approx x \):
\[ \tan\left(\frac{\pi}{4} + x\right) \approx \frac{1 + x}{1 - x} \]

### Step 4: Simplify the expression
Now we can simplify:
\[ \tan\left(\frac{\pi}{4} + x\right) \approx \frac{1 + x}{1 - x} = \frac{(1 + x)(1 + x)}{(1 - x)(1 + x)} = \frac{1 + 2x + x^2}{1 - x^2} \]
As \( x \to 0 \), this approaches \( 1 + 2x \).

### Step 5: Substitute back into the limit
Now substituting back into our limit:
\[ L = \lim_{x \to 0} \left(1 + 2x\right)^{\frac{1}{x}} \]

### Step 6: Recognize the indeterminate form
This limit is of the form \( 1^{\infty} \), so we can use the exponential limit property:
\[ L = e^{\lim_{x \to 0} \frac{(1 + 2x - 1)}{x}} = e^{\lim_{x \to 0} \frac{2x}{x}} = e^{\lim_{x \to 0} 2} = e^2 \]

### Final Answer
Thus, the limit is:
\[ L = e^2 \]