To solve the limit:

$\lim_{x \to 0} \frac{\sin(3x) \cdot \cot(5x)}{x \cdot \cot(4x)},$

we proceed as follows:

Step-by-Step Solution:

1. Simplify the trigonometric expressions:

• Recall that $\cot(y) = \frac{\cos(y)}{\sin(y)}$. So, $\cot(5x) = \frac{\cos(5x)}{\sin(5x)}, \quad \cot(4x) = \frac{\cos(4x)}{\sin(4x)}.$

Substituting these into the limit: $\frac{\sin(3x) \cdot \frac{\cos(5x)}{\sin(5x)}}{x \cdot \frac{\cos(4x)}{\sin(4x)}} = \frac{\sin(3x) \cos(5x) \sin(4x)}{x \sin(5x) \cos(4x)}.$

2. Apply trigonometric approximations for small $x$:

• As $x \to 0$:
  • $\sin(kx) \sim kx$,
  • $\cos(kx) \sim 1$.

Using these approximations:

• $\sin(3x) \sim 3x$,
• $\sin(5x) \sim 5x$,
• $\sin(4x) \sim 4x$,
• $\cos(5x) \sim 1$,
• $\cos(4x) \sim 1$.

Substituting these approximations: $\frac{(3x)(1)(4x)}{x (5x)(1)} = \frac{12x^2}{5x^2}.$

3. Simplify the expression:

Cancel $x^2$ in the numerator and denominator: $\frac{12}{5}.$

Final Answer:

$\lim_{x \to 0} \frac{\sin(3x) \cot(5x)}{x \cot(4x)} = \frac{12}{5}.$

Would you like further clarification or help with related problems? Here are some relative questions you might find useful:

1. What is the importance of small-angle approximations in limits involving trigonometric functions?
2. How does L'Hôpital's Rule apply to limits like this one?
3. Can this limit be evaluated using series expansion instead?
4. What are common errors to watch for when simplifying trigonometric limits?
5. How does this problem differ when approaching infinity instead of zero?

Tip: Always double-check small-angle approxim