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

Explanation

To find 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: Identify the Form
First, we substitute \( x = 0 \):
\[ \tan\left(\frac{\pi}{4} + 0\right) = \tan\left(\frac{\pi}{4}\right) = 1 \]
Thus, the expression becomes \( 1^{\infty} \), which is an indeterminate form.

### Step 2: Rewrite the Limit
We can rewrite the limit using the exponential function:
\[ y = \left( \tan\left(\frac{\pi}{4} + x\right) \right)^{\frac{1}{x}} \implies \ln y = \frac{1}{x} \ln\left( \tan\left(\frac{\pi}{4} + x\right) \right) \]
Now we need to find:
\[ \lim_{x \to 0} \ln y = \lim_{x \to 0} \frac{\ln\left( \tan\left(\frac{\pi}{4} + x\right) \right)}{x} \]

### Step 3: Apply L'Hôpital's Rule
Since substituting \( x = 0 \) gives us \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{\ln\left( \tan\left(\frac{\pi}{4} + x\right) \right)}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}\left(\ln\left( \tan\left(\frac{\pi}{4} + x\right)\right)\right)}{\frac{d}{dx}(x)} \]

### Step 4: Differentiate the Numerator
Using the chain rule, we differentiate:
\[ \frac{d}{dx}\left(\ln\left( \tan\left(\frac{\pi}{4} + x\right)\right)\right) = \frac{1}{\tan\left(\frac{\pi}{4} + x\right)} \cdot \sec^2\left(\frac{\pi}{4} + x\right) \]
Thus, we have:
\[ \lim_{x \to 0} \frac{\sec^2\left(\frac{\pi}{4} + x\right)}{\tan\left(\frac{\pi}{4} + x\right)} \]

### Step 5: Evaluate the Limit
Now substituting \( x = 0 \):
\[ \sec^2\left(\frac{\pi}{4}\right) = 2 \quad \text{and} \quad \tan\left(\frac{\pi}{4}\right) = 1 \]
So the limit becomes:
\[ \lim_{x \to 0} \frac{2}{1} = 2 \]

### Step 6: Exponentiate to Find \( y \)
Now we have:
\[ \ln y = 2 \implies y = e^2 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \left( \tan\left(\frac{\pi}{4} + x\right) \right)^{\frac{1}{x}} = e^2 \]

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