We are tasked with evaluating the following limit:

$\lim_{x \to 0} \frac{C \cdot \tan^2(x) \cdot \csc(x)}{x}$

where $C$ is a nonzero real number.

Step-by-step evaluation:

1. Break down the expression: The given expression can be rewritten as: $\lim_{x \to 0} \frac{C \cdot \tan^2(x) \cdot \csc(x)}{x} = C \cdot \lim_{x \to 0} \frac{\tan^2(x) \cdot \csc(x)}{x}$ Since $C$ is a constant, we can focus on evaluating the limit of the function $\frac{\tan^2(x) \cdot \csc(x)}{x}$.

2. Recall important approximations near zero: As $x \to 0$, we can use the following approximations:

   • $\tan(x) \approx x$, meaning $\tan^2(x) \approx x^2$.
   • $\csc(x) = \frac{1}{\sin(x)} \approx \frac{1}{x}$.

   Using these, we approximate the expression: $\frac{\tan^2(x) \cdot \csc(x)}{x} \approx \frac{x^2 \cdot \frac{1}{x}}{x} = \frac{x^2}{x^2} = 1.$

3. Final answer: Since the limit of $\frac{\tan^2(x) \cdot \csc(x)}{x}$ approaches 1 as $x \to 0$, the overall limit is: $C \cdot 1 = C.$

Thus, the value of the limit is:

$\boxed{C}.$

Would you like more details or have further questions? Here are some related questions for further exploration:

1. How would this limit change if $C = 0$?
2. Why do $\tan(x)$ and $\sin(x)$ have similar behaviors near zero?
3. How do small angle approximations help in calculating limits?
4. Can you calculate the limit without using approximations?
5. What other limits involving trigonometric functions are useful?

Tip: Small angle approximations are very useful for evaluating limits of trigonometric functions near zero.