We are tasked with evaluating the limit:

$\lim_{x \to 0} \frac{1}{3x \cdot \csc(x)}.$

Step-by-step solution:

1. Recall the definition of $\csc(x)$: $\csc(x) = \frac{1}{\sin(x)}.$ Substituting this into the expression, we get: $\frac{1}{3x \cdot \csc(x)} = \frac{1}{3x \cdot \frac{1}{\sin(x)}} = \frac{\sin(x)}{3x}.$

2. Simplify the limit: The expression reduces to: $\lim_{x \to 0} \frac{\sin(x)}{3x}.$

3. Use the standard limit property: $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$. Thus: $\lim_{x \to 0} \frac{\sin(x)}{3x} = \frac{1}{3}.$

Final Answer:

$\boxed{\frac{1}{3}}$

Let me know if you want further clarification or deeper insights into this problem!

Here are 5 related questions to deepen your understanding:

1. What is the limit of $\frac{\tan(x)}{x}$ as $x \to 0$?
2. How would the answer change if the expression was $\lim_{x \to 0} \frac{1}{2x \cdot \csc(x)}$?
3. What is the significance of the standard limit $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ in calculus?
4. How would you evaluate the limit if $\sin(x)$ were replaced by $\sin^2(x)$?
5. Can L’Hôpital’s Rule be used here? Why or why not?

Tip: Familiarize yourself with standard trigonometric limits; they frequently appear in calculus problems.