The problem asks us to evaluate the following limit: $ \lim_{x\to\frac{\pi}{3}} \frac{\sqrt{3}(\frac{\pi}{3}-x)}{\sin(x-\frac{\pi}{3})} $

1. Problem Description

The problem asks us to evaluate the following limit:

$\lim_{x\to\frac{\pi}{3}} \frac{\sqrt{3}(\frac{\pi}{3}-x)}{\sin(x-\frac{\pi}{3})}$

2. Solution Steps

Let $u = x - \frac{\pi}{3}$. Then $x = u + \frac{\pi}{3}$.

As $x \to \frac{\pi}{3}$, we have $u \to 0$.

Then, $\frac{\pi}{3} - x = \frac{\pi}{3} - (u+\frac{\pi}{3}) = -u$.

Thus, we have

$$$$

\lim_{x\to\frac{\pi}{3}} \frac{\sqrt{3}(\frac{\pi}{3}-x)}{\sin(x-\frac{\pi}{3})} = \lim_{u\to 0} \frac{\sqrt{3}(-u)}{\sin(u)} = -\sqrt{3} \lim_{u\to 0} \frac{u}{\sin(u)}

$$$$

We know that

$$$$

\lim_{x\to 0} \frac{\sin x}{x} = 1

$$$$

Therefore,

$$$$

\lim_{x\to 0} \frac{x}{\sin x} = \frac{1}{\lim_{x\to 0} \frac{\sin x}{x}} = \frac{1}{1} = 1

$$$$

So,

$$$$

-\sqrt{3} \lim_{u\to 0} \frac{u}{\sin(u)} = -\sqrt{3} \times 1 = -\sqrt{3}

$$$$

3. Final Answer

$-\sqrt{3}$