The value of `lim_(xto1)(tan((pi)/4+logx))^(1/(logx))` is equal to

Options

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

Answer

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

### Step 1: Identify the Limit Form
First, we substitute \( x = 1 \) into the expression:
\[ \tan\left(\frac{\pi}{4} + \log(1)\right)^{\frac{1}{\log(1)}} \]
Since \( \log(1) = 0 \), this results in:
\[ \tan\left(\frac{\pi}{4} + 0\right)^{\frac{1}{0}} = \tan\left(\frac{\pi}{4}\right)^{\infty} = 1^{\infty} \]
This is an indeterminate form of type \( 1^{\infty} \).

### Step 2: Rewrite the Limit
We can use the standard limit result for \( a^{b} \) where \( a \to 1 \) and \( b \to \infty \):
\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} g(x) (f(x) - 1)} \]
Here, let \( f(x) = \tan\left(\frac{\pi}{4} + \log x\right) \) and \( g(x) = \frac{1}{\log x} \).

### Step 3: Calculate \( f(x) - 1 \)
We need to find:
\[ f(x) - 1 = \tan\left(\frac{\pi}{4} + \log x\right) - 1 \]
Using the identity \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \), where \( A = \frac{\pi}{4} \) and \( B = \log x \):
\[ \tan\left(\frac{\pi}{4} + \log x\right) = \frac{1 + \tan(\log x)}{1 - \tan(\log x)} \]
Thus,
\[ f(x) - 1 = \frac{1 + \tan(\log x)}{1 - \tan(\log x)} - 1 = \frac{1 + \tan(\log x) - (1 - \tan(\log x))}{1 - \tan(\log x)} = \frac{2\tan(\log x)}{1 - \tan(\log x)} \]

### Step 4: Evaluate the Limit
Now, we need to evaluate:
\[ \lim_{x \to 1} \frac{1}{\log x} \cdot \frac{2\tan(\log x)}{1 - \tan(\log x)} \]
As \( x \to 1 \), \( \log x \to 0 \), and thus \( \tan(\log x) \to \tan(0) = 0 \). This gives us a \( \frac{0}{0} \) form.

### Step 5: Apply L'Hôpital's Rule
We differentiate the numerator and denominator:
- The derivative of \( 2\tan(\log x) \) is \( 2\sec^2(\log x) \cdot \frac{1}{x} \).
- The derivative of \( 1 - \tan(\log x) \) is \( -\sec^2(\log x) \cdot \frac{1}{x} \).

Thus, applying L'Hôpital's Rule:
\[ \lim_{x \to 1} \frac{2\sec^2(\log x) \cdot \frac{1}{x}}{-\sec^2(\log x) \cdot \frac{1}{x}} = \lim_{x \to 1} -2 = -2 \]

### Step 6: Final Result
Now substituting back into the limit:
\[ \lim_{x \to 1} f(x)^{g(x)} = e^{-2} \]

Thus, the value of the limit is:
\[ \lim_{x \to 1} \left( \tan\left(\frac{\pi}{4} + \log x\right) \right)^{\frac{1}{\log x}} = e^{-2} \]