The limit expression extracted from the image seems to be:

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

Let me now proceed to solve this limit step by step.

We begin by substituting $x = \frac{\pi}{3}$ into the numerator and denominator.

Step 1: Substitution

Numerator: $\sin\left(\frac{\pi}{3}\right) - \sqrt{3} \cos\left(\frac{\pi}{3}\right)$ Denominator: $\sin\left( 3 \times \frac{\pi}{3} \right) = \sin(\pi)$

Since $\sin(\pi) = 0$, we have an indeterminate form $\frac{0}{0}$, meaning L'Hôpital's rule applies.

Step 2: L'Hôpital's Rule

We differentiate the numerator and denominator:

Numerator: $\frac{d}{dx} \left( \sin{x} - \sqrt{3} \cos{x} \right) = \cos{x} + \sqrt{3} \sin{x}$

Denominator: $\frac{d}{dx} \left( \sin{3x} \right) = 3 \cos{3x}$

Step 3: Evaluate the new limit

Now, the limit becomes: $\lim_{x \to \frac{\pi}{3}} \frac{\cos{x} + \sqrt{3} \sin{x}}{3 \cos{3x}}$

Substitute $x = \frac{\pi}{3}$ again:

• $\cos\left( \frac{\pi}{3} \right) = \frac{1}{2}$
• $\sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}$
• $\cos(3 \times \frac{\pi}{3}) = \cos(\pi) = -1$

Thus, the limit becomes: $\frac{\frac{1}{2} + \sqrt{3} \times \frac{\sqrt{3}}{2}}{3 \times (-1)} = \frac{\frac{1}{2} + \frac{3}{2}}{-3} = \frac{2}{-3} = -\frac{2}{3}$

Final Answer:

$\lim_{x \to \frac{\pi}{3}} \frac{\sin{x} - \sqrt{3} \cos{x}}{\sin{3x}} = -\frac{2}{3}$

Would you like further details on the steps or explanations? Here are five related questions you can explore:

1. How does L'Hôpital's rule apply to other indeterminate forms like $\frac{\infty}{\infty}$?
2. What are the common indeterminate forms that appear in limit problems?
3. Can we apply a Taylor series expansion to solve limits at points other than 0?
4. What are other techniques for evaluating trigonometric limits without using L'Hôpital's rule?
5. How do derivatives of trigonometric functions help in solving limit problems?

Tip: Always check for indeterminate forms before applying L'Hôpital's rule to ensure it's valid.