We are given the limit expression:

$A = \lim_{x \to 0} \left( (1 - 8\sin(x))^{-8} + \frac{3x}{\sin(x)} \right)$

Step 1: Analyze the first part: $(1 - 8\sin(x))^{-8}$

As $x \to 0$, we know that $\sin(x) \approx x$, so:

$1 - 8\sin(x) \approx 1 - 8x$

Thus, for small $x$, we can approximate:

$(1 - 8\sin(x))^{-8} \approx (1 - 8x)^{-8}$

Using the binomial expansion for $(1 - 8x)^{-8}$ when $x$ is small:

$(1 - 8x)^{-8} = 1 + 8 \cdot 8x + \mathcal{O}(x^2) = 1 + 64x + \mathcal{O}(x^2)$

Therefore, as $x \to 0$:

$(1 - 8\sin(x))^{-8} \approx 1 + 64x + \mathcal{O}(x^2)$

Step 2: Analyze the second part: $\frac{3x}{\sin(x)}$

As $x \to 0$, we know the standard limit:

$\frac{\sin(x)}{x} \to 1 \quad \text{so} \quad \frac{3x}{\sin(x)} \to 3$

Thus:

$\frac{3x}{\sin(x)} \approx 3 + \mathcal{O}(x^2)$

Step 3: Combine the results

Now, combining the results from both parts:

$A = \lim_{x \to 0} \left( (1 - 8\sin(x))^{-8} + \frac{3x}{\sin(x)} \right)$

$A = \lim_{x \to 0} \left( 1 + 64x + \mathcal{O}(x^2) + 3 + \mathcal{O}(x^2) \right)$

$A = 1 + 3 + 64x + \mathcal{O}(x^2) = 4 + \mathcal{O}(x)$

As $x \to 0$, the $\mathcal{O}(x)$ term vanishes, and we are left with:

$A = 4$

Final Answer:

$A = 4$

Would you like a more detailed breakdown of any step, or is there anything else you'd like to ask about this? Here are some follow-up questions to deepen your understanding:

1. How does the approximation $\sin(x) \approx x$ help in simplifying limits?
2. Can we use a Taylor series expansion to solve limits like this?
3. What does the $\mathcal{O}(x^2)$ term mean in the context of small $x$?
4. How does the binomial expansion apply to small $x$?
5. Why does the term involving $64x$ vanish as $x \to 0$?

Tip: For limits involving small values of $x$, it's often helpful to use approximations like $\sin(x) \approx x$, and expand functions using series to capture the behavior near zero.