`lim_( x to pi//6) (sin(x-pi//6))/(sqrt3//2-cosx)` is equal to

Options

A. 2
B. 1
C. `-1`
D. `sqrt3//2`

Answer

To solve the limit \( \lim_{x \to \frac{\pi}{6}} \frac{\sin(x - \frac{\pi}{6})}{\frac{\sqrt{3}}{2} - \cos x} \), we can follow these steps:

### Step 1: Substitute \( x = \frac{\pi}{6} \)
First, we substitute \( x = \frac{\pi}{6} \) directly into the expression.

\[ \sin\left(\frac{\pi}{6} - \frac{\pi}{6}\right) = \sin(0) = 0 \]
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
Thus, the denominator becomes:
\[ \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = 0 \]

This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator.

### Step 3: Differentiate the Numerator and Denominator
The numerator is \( \sin(x - \frac{\pi}{6}) \) and the denominator is \( \frac{\sqrt{3}}{2} - \cos x \).

1. Differentiate the numerator:
\[ \frac{d}{dx}[\sin(x - \frac{\pi}{6})] = \cos(x - \frac{\pi}{6}) \cdot \frac{d}{dx}(x - \frac{\pi}{6}) = \cos(x - \frac{\pi}{6}) \]

2. Differentiate the denominator:
\[ \frac{d}{dx}\left[\frac{\sqrt{3}}{2} - \cos x\right] = \sin x \]

### Step 4: Rewrite the Limit
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to \frac{\pi}{6}} \frac{\cos(x - \frac{\pi}{6})}{\sin x} \]

### Step 5: Substitute \( x = \frac{\pi}{6} \) Again
Now we substitute \( x = \frac{\pi}{6} \) into the new limit:
1. For the numerator:
\[ \cos\left(\frac{\pi}{6} - \frac{\pi}{6}\right) = \cos(0) = 1 \]

2. For the denominator:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

### Step 6: Calculate the Limit
Now we can compute the limit:
\[ \lim_{x \to \frac{\pi}{6}} \frac{1}{\frac{1}{2}} = 2 \]

### Final Answer
Thus, the limit is:
\[ \boxed{2} \]