The limit you provided is:

$\lim_{x \to 0} \frac{\cos 4x - 1 + \sin 6x}{\tan 2x}$

To solve this limit, let's start by analyzing the function as $x$ approaches 0. First, we use the Taylor series expansions for the trigonometric functions around $x = 0$:

1. Cosine Expansion: $\cos(4x) \approx 1 - \frac{(4x)^2}{2} = 1 - 8x^2 + \mathcal{O}(x^4)$

2. Sine Expansion: $\sin(6x) \approx 6x - \frac{(6x)^3}{6} = 6x - 36x^3 + \mathcal{O}(x^5)$

3. Tangent Expansion: $\tan(2x) \approx 2x + \frac{(2x)^3}{3} = 2x + \frac{8x^3}{3} + \mathcal{O}(x^5)$

Substitute these expansions into the original limit:

$\lim_{x \to 0} \frac{(1 - 8x^2 - 1 + 6x - 36x^3)}{2x + \frac{8x^3}{3}}$

Simplify the expression:

$\lim_{x \to 0} \frac{-8x^2 + 6x - 36x^3}{2x + \frac{8x^3}{3}}$

Now, factor out $x$ from both the numerator and the denominator:

$\lim_{x \to 0} \frac{x(-8x + 6 - 36x^2)}{x(2 + \frac{8x^2}{3})}$

This simplifies to:

$\lim_{x \to 0} \frac{-8x + 6 - 36x^2}{2 + \frac{8x^2}{3}}$

Now, evaluate the limit as $x$ approaches 0:

$\frac{6}{2} = 3$

So, the limit is:

$\boxed{3}$

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How do Taylor series expansions help in evaluating limits?
2. What are other methods for evaluating limits that involve indeterminate forms?
3. Can L'Hôpital's Rule be applied to this problem? Why or why not?
4. How do small-angle approximations for sine and cosine functions relate to Taylor expansions?
5. What is the geometric interpretation of the limit as $x$ approaches 0?

Tip: When dealing with trigonometric limits, knowing the small-angle approximations and Taylor series expansions can simplify the process significantly.