`underset(x to 0^(+))lim log_(tan x) (tan 2x)` is equal to

Options

A. 1
B. `-1
C. 0
D. None of these

Answer

To solve the limit \( \lim_{x \to 0^+} \log_{\tan x} (\tan 2x) \), we can follow these steps:

### Step 1: Rewrite the logarithm
We can rewrite the logarithm in terms of natural logarithms (base \( e \)):
\[ \log_{\tan x} (\tan 2x) = \frac{\log(\tan 2x)}{\log(\tan x)} \]
Thus, the limit becomes:
\[ \lim_{x \to 0^+} \frac{\log(\tan 2x)}{\log(\tan x)} \]

### Step 2: Analyze the limit
As \( x \to 0^+ \), both \( \tan 2x \) and \( \tan x \) approach 0. Therefore, \( \log(\tan 2x) \) and \( \log(\tan x) \) both approach \( -\infty \). This gives us the indeterminate form \( \frac{-\infty}{-\infty} \).

### Step 3: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
if the limit on the right side exists. We differentiate the numerator and denominator:
- The derivative of \( \log(\tan 2x) \) is:
\[ \frac{d}{dx}[\log(\tan 2x)] = \frac{1}{\tan 2x} \cdot \sec^2(2x) \cdot 2 = \frac{2 \sec^2(2x)}{\tan 2x} \]
- The derivative of \( \log(\tan x) \) is:
\[ \frac{d}{dx}[\log(\tan x)] = \frac{1}{\tan x} \cdot \sec^2(x) = \frac{\sec^2(x)}{\tan x} \]

### Step 4: Rewrite the limit using derivatives
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 0^+} \frac{\frac{2 \sec^2(2x)}{\tan 2x}}{\frac{\sec^2(x)}{\tan x}} = \lim_{x \to 0^+} \frac{2 \sec^2(2x) \tan x}{\sec^2(x) \tan 2x} \]

### Step 5: Evaluate the limit
As \( x \to 0^+ \):
- \( \tan x \approx x \)
- \( \tan 2x \approx 2x \)
- \( \sec^2(2x) \to 1 \)
- \( \sec^2(x) \to 1 \)

Thus, the limit simplifies to:
\[ \lim_{x \to 0^+} \frac{2 \cdot 1 \cdot x}{1 \cdot 2x} = \lim_{x \to 0^+} \frac{2x}{2x} = 1 \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to 0^+} \log_{\tan x} (\tan 2x) = 1 \]

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