`lim_(xrarr0)(sinx)^(2tanx)`

Options

A. is 2
B. is 1
C. is 0
D. does not exist

Answer

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

### Step 1: Define the limit
Let \( y = (\sin x)^{2 \tan x} \).

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides, we have:
\[ \ln y = 2 \tan x \cdot \ln(\sin x) \]

### Step 3: Analyze the limit
We need to find:
\[ \lim_{x \to 0} \ln y = \lim_{x \to 0} 2 \tan x \cdot \ln(\sin x) \]
As \( x \to 0 \), both \( \tan x \) and \( \ln(\sin x) \) approach 0, leading to an indeterminate form \( 0 \cdot (-\infty) \).

### Step 4: Rewrite the limit
To resolve this, we can rewrite the limit as:
\[ \lim_{x \to 0} 2 \cdot \frac{\ln(\sin x)}{\cot x} \]
This is now in the form \( \frac{0}{0} \).

### Step 5: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{\ln(\sin x)}{\cot x} = \lim_{x \to 0} \frac{\frac{d}{dx}(\ln(\sin x))}{\frac{d}{dx}(\cot x)} \]

### Step 6: Differentiate the numerator and denominator
The derivative of \( \ln(\sin x) \) is:
\[ \frac{1}{\sin x} \cdot \cos x = \cot x \]
The derivative of \( \cot x \) is:
\[ -\csc^2 x \]
Thus, we have:
\[ \lim_{x \to 0} \frac{\cot x}{-\csc^2 x} = \lim_{x \to 0} \frac{\cot x \cdot \sin^2 x}{-1} \]

### Step 7: Simplify the limit
This simplifies to:
\[ \lim_{x \to 0} -\sin^2 x \cdot \cot x = \lim_{x \to 0} -\sin^2 x \cdot \frac{\cos x}{\sin x} = \lim_{x \to 0} -\sin x \cdot \cos x \]
As \( x \to 0 \), this limit approaches:
\[ 0 \]

### Step 8: Conclude the limit for \( \ln y \)
Thus, we have:
\[ \lim_{x \to 0} \ln y = 0 \]

### Step 9: Exponentiate to find \( y \)
Exponentiating both sides gives:
\[ y = e^0 = 1 \]

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

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