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The problem asks us to compute the limit:

\[
L = \lim_{x \to 0} \frac{3 \arcsin^3 x - 3 \arcsin^2 x - 4 \arcsin x}{5x^3 - 6x^2 + 3x}
\]

To solve this, let’s follow these steps:

### Step 1: Series Expansion of \(\arcsin x\)

Since \(\arcsin x\) is a standard elementary function, we can expand it in a Taylor series around \(x = 0\). The series expansion of \(\arcsin x\) near \(x = 0\) is:

\[
\arcsin x = x + \frac{x^3}{6} + O(x^5)
\]

### Step 2: Substitution of \(\arcsin x\)

Now, we substitute this expansion into each term of the numerator.

1. \(\arcsin^3 x = \left(x + \frac{x^3}{6} + O(x^5)\right)^3 = x^3 + O(x^5)\)
2. \(\arcsin^2 x = \left(x + \frac{x^3}{6} + O(x^5)\right)^2 = x^2 + O(x^4)\)
3. \(\arcsin x = x + O(x^3)\)

So, the numerator becomes:
\[
3 \arcsin^3 x - 3 \arcsin^2 x - 4 \arcsin x = 3(x^3 + O(x^5)) - 3(x^2 + O(x^4)) - 4(x + O(x^3))
\]
Simplifying:
\[
= 3x^3 - 3x^2 - 4x + O(x^4)
\]

### Step 3: Simplifying the Denominator

The denominator is:
\[
5x^3 - 6x^2 + 3x
\]
This is already in simplified form.

### Step 4: Forming the Limit Expression

Now, we substitute both the numerator and denominator into the limit expression:

\[
\frac{3x^3 - 3x^2 - 4x + O(x^4)}{5x^3 - 6x^2 + 3x}
\]

### Step 5: Factorizing and Canceling Common Terms

We can factor \(x\) from both the numerator and denominator:

\[
\frac{x(3x^2 - 3x - 4 + O(x^3))}{x(5x^2 - 6x + 3)}
\]

Canceling the common \(x\):

\[
\frac{3x^2 - 3x - 4 + O(x^3)}{5x^2 - 6x + 3}
\]

### Step 6: Evaluating the Limit as \(x \to 0\)

As \(x \to 0\), the higher-order terms (\(O(x^3)\)) vanish, and we are left with:

\[
\frac{-4}{3}
\]

Thus, the value of the limit is:

\[
L = -\frac{4}{3}
\]

The correct answer is \(\boxed{B}\).

---

Would you like further clarification on any steps? Here are some related questions to consider:

1. How is the Taylor series expansion of \(\arcsin x\) derived?
2. Can you provide more examples of limits involving trigonometric functions?
3. How do we handle indeterminate forms like \( \frac{0}{0} \) in limits?
4. Why does the term \(O(x^n)\) vanish as \(x \to 0\)?
5. Can this limit be solved using L'Hopital's Rule?

**Tip**: Always check if a Taylor series expansion can simplify limits, especially when dealing with trigonometric or logarithmic functions near 0.

Compute the limit: L = lim(x→0) [(3 * arcsin^3(x) - 3 * arcsin^2(x) - 4 * arcsin(x)) / (5x^3 - 6x^2 + 3x)]

Learn how to compute the limit lim(x→0) [(3 * arcsin^3(x) - 3 * arcsin^2(x) - 4 * arcsin(x)) / (5x^3 - 6x^2 + 3x)] using the Taylor series expansion. This detailed solution covers series expansion techniques and limit evaluation.

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