Evaluate `underset(x to pi//6)lim (2sin x-1)/(x-pi/6)`

Options

A. 3
B. `sqrt3`
C. 1
D. `-1`

Answer

To evaluate the limit

\[ \lim_{x \to \frac{\pi}{6}} \frac{2 \sin x - 1}{x - \frac{\pi}{6}}, \]

we will follow these steps:

### Step 1: Substitute the limit value
First, we substitute \( x = \frac{\pi}{6} \) into the expression to check the form of the limit.

\[ 2 \sin\left(\frac{\pi}{6}\right) - 1 = 2 \cdot \frac{1}{2} - 1 = 1 - 1 = 0, \]
and
\[ x - \frac{\pi}{6} = \frac{\pi}{6} - \frac{\pi}{6} = 0. \]

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

### Step 2: Apply L'Hôpital's Rule
Since we have the indeterminate form \( \frac{0}{0} \), 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.

Here, let \( f(x) = 2 \sin x - 1 \) and \( g(x) = x - \frac{\pi}{6} \).

### Step 3: Differentiate the numerator and denominator
Now we differentiate \( f(x) \) and \( g(x) \):

- The derivative of \( f(x) = 2 \sin x \) is \( f'(x) = 2 \cos x \).
- The derivative of \( g(x) = x - \frac{\pi}{6} \) is \( g'(x) = 1 \).

### Step 4: Rewrite the limit using derivatives
Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to \frac{\pi}{6}} \frac{2 \sin x - 1}{x - \frac{\pi}{6}} = \lim_{x \to \frac{\pi}{6}} \frac{2 \cos x}{1}. \]

### Step 5: Substitute \( x = \frac{\pi}{6} \) again
Now we substitute \( x = \frac{\pi}{6} \) into the new limit:

\[ 2 \cos\left(\frac{\pi}{6}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}. \]

### Final Answer
Thus, the limit evaluates to:

\[ \lim_{x \to \frac{\pi}{6}} \frac{2 \sin x - 1}{x - \frac{\pi}{6}} = \sqrt{3}. \]