To solve this limit problem:

$\lim_{x \to 0} \frac{4x \cdot \tan(3x)}{1 - \cos(4x)}$

we'll use limit properties and trigonometric approximations for small $x$. Here is the detailed solution:

Step 1: Expand Trigonometric Functions

For small values of $x$, we use the following approximations:

• $\tan(3x) \approx 3x$
• $1 - \cos(4x) \approx \frac{(4x)^2}{2} = 8x^2$

Substitute these approximations into the limit:

$\lim_{x \to 0} \frac{4x \cdot 3x}{8x^2}$

Step 2: Simplify the Expression

Simplify the above expression:

$\lim_{x \to 0} \frac{12x^2}{8x^2}$

The $x^2$ terms cancel out:

$\lim_{x \to 0} \frac{12}{8} = \frac{3}{2}$

Final Answer

The result of the limit is $\frac{3}{2}$, which corresponds to option (c).

Conclusion

Thus, the correct answer is:

c. $\frac{3}{2}$

Would you like further details or have any other questions?

Here are 5 related questions that expand on the topic:

1. How do you derive the small-angle approximations for trigonometric functions?
2. What other methods can be used to solve limits involving trigonometric functions?
3. How can L'Hôpital's rule be applied to solve this type of limit?
4. Why does $\tan(x)$ approximate to $x$ as $x \to 0$?
5. What are some common pitfalls when solving trigonometric limits?

Tip: When solving limits involving trigonometric functions, remember that small-angle approximations can simplify the calculations significantly, especially when $x$ approaches 0.