The problem you uploaded asks to find the following limit:

$\lim_{x \to 0} \frac{\cos(4x) \cdot \tan(4x)}{6x}$

Step-by-Step Solution:

1. Use Taylor Expansion or Standard Limits: Near $x = 0$, we can use approximations for small angles. Recall that:

   • $\cos(4x) \approx 1$ for small $x$,
   • $\tan(4x) \approx 4x$ for small $x$ (since $\tan(x) \approx x$ when $x \to 0$).

2. Simplify the Expression: Substituting these approximations into the limit:

$\frac{\cos(4x) \cdot \tan(4x)}{6x} \approx \frac{1 \cdot 4x}{6x} = \frac{4x}{6x} = \frac{4}{6} = \frac{2}{3}$

3. Conclusion: The value of the limit is:

$\boxed{\frac{2}{3}}$

Let me know if you would like more details or have any further questions!

Related Questions:

1. How does the approximation $\tan x \approx x$ work for small angles?
2. Can we solve this limit using L'Hopital's Rule? How?
3. What are some other common trigonometric limits near $x = 0$?
4. How do Taylor series expansions help in solving limits involving trigonometric functions?
5. How does the result change if the angle inside the trigonometric functions is different (e.g., $2x$ instead of $4x$)?

Tip: When dealing with trigonometric limits, small-angle approximations and standard limit rules like $\lim_{x \to 0} \frac{\sin x}{x} = 1$ are often very useful!