To calculate the given limit using L'Hôpital's Rule, we analyze it step by step. The given limit is:

$\lim_{x \to 0} \frac{\arcsin(2x) - 2\arcsin(x)}{x^3}.$

Step 1: Check indeterminate form

As $x \to 0$,

• $\arcsin(2x) \to \arcsin(0) = 0$,
• $\arcsin(x) \to \arcsin(0) = 0$.

So, the numerator becomes $0 - 0 = 0$ and the denominator is $x^3 \to 0$. Hence, the limit is in the indeterminate form $0/0$, and we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

Take the derivative of the numerator and denominator:

• Numerator: Derivative of $\arcsin(2x)$ is $\frac{d}{dx}[\arcsin(2x)] = \frac{2}{\sqrt{1-(2x)^2}}$, and derivative of $-2\arcsin(x)$ is $-\frac{2}{\sqrt{1-x^2}}$. So, the derivative of the numerator is: $\frac{2}{\sqrt{1-(2x)^2}} - \frac{2}{\sqrt{1-x^2}}.$

• Denominator: Derivative of $x^3$ is $3x^2$.

Thus, the limit becomes: $\lim_{x \to 0} \frac{\frac{2}{\sqrt{1-(2x)^2}} - \frac{2}{\sqrt{1-x^2}}}{3x^2}.$

Step 3: Simplify and reapply L'Hôpital's Rule

At $x = 0$, the numerator is still $0$ and the denominator is $0$. We apply L'Hôpital's Rule again.

Derivative of the numerator:

Using the chain rule, compute the derivatives of: $\frac{d}{dx}\left(\frac{2}{\sqrt{1-(2x)^2}}\right) = \frac{2 \cdot (4x)}{(1-(2x)^2)^{3/2}}$ and $\frac{d}{dx}\left(-\frac{2}{\sqrt{1-x^2}}\right) = \frac{2 \cdot x}{(1-x^2)^{3/2}}.$

So, the derivative of the numerator is: $\frac{8x}{(1-(2x)^2)^{3/2}} - \frac{2x}{(1-x^2)^{3/2}}.$

Derivative of the denominator:

$\frac{d}{dx}[3x^2] = 6x.$

The new limit becomes: $\lim_{x \to 0} \frac{\frac{8x}{(1-(2x)^2)^{3/2}} - \frac{2x}{(1-x^2)^{3/2}}}{6x}.$

Step 4: Simplify further

Factor $x$ out of the numerator and cancel with the $x$ in the denominator: $\lim_{x \to 0} \frac{\frac{8}{(1-(2x)^2)^{3/2}} - \frac{2}{(1-x^2)^{3/2}}}{6}.$

As $x \to 0$, both $1-(2x)^2$ and $1-x^2$ approach $1$, so: $\frac{8}{(1-(2x)^2)^{3/2}} \to 8, \quad \frac{2}{(1-x^2)^{3/2}} \to 2.$

Thus, the numerator becomes: $8 - 2 = 6.$

The denominator is $6$. Therefore, the limit is: $\frac{6}{6} = 1.$

Final Answer:

$\boxed{1}$

Would you like a deeper explanation of any step or related examples?

Related Questions:

1. What are the conditions for applying L'Hôpital's Rule?
2. How do trigonometric limits like $\arcsin(x)$ behave near zero?
3. Can this problem be solved without L'Hôpital’s Rule?
4. What other indeterminate forms are suitable for L'Hôpital's Rule?
5. How does the chain rule affect derivatives of composite functions like $\arcsin(2x)$?

Tip: Always verify the indeterminate form before applying L'Hôpital's Rule.