`lim_(xto0)(sin^(-1)x-tan^(-1)x)/(x^(3))` is equal to

Options

A. 0
B. 1
C. -1
D. `1/2`

Explanation

To solve the limit

\[ \lim_{x \to 0} \frac{\sin^{-1} x - \tan^{-1} x}{x^3}, \]

we first notice that substituting \( x = 0 \) directly into the expression gives us the indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and the denominator until we reach a determinate form.

### Step 1: Differentiate the numerator and the denominator

The numerator is \( \sin^{-1} x - \tan^{-1} x \). We differentiate it:

- The derivative of \( \sin^{-1} x \) is \( \frac{1}{\sqrt{1 - x^2}} \).
- The derivative of \( \tan^{-1} x \) is \( \frac{1}{1 + x^2} \).

Thus, the derivative of the numerator is:

\[ \frac{d}{dx}(\sin^{-1} x - \tan^{-1} x) = \frac{1}{\sqrt{1 - x^2}} - \frac{1}{1 + x^2}. \]

The denominator \( x^3 \) differentiates to \( 3x^2 \).

Now we can rewrite our limit as:

\[ \lim_{x \to 0} \frac{\frac{1}{\sqrt{1 - x^2}} - \frac{1}{1 + x^2}}{3x^2}. \]

### Step 2: Substitute \( x = 0 \) again

Substituting \( x = 0 \) into the new expression still gives us \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 3: Differentiate again

We need to differentiate the new numerator and the denominator again.

1. Differentiate the numerator:

The numerator is \( \frac{1}{\sqrt{1 - x^2}} - \frac{1}{1 + x^2} \).

- The derivative of \( \frac{1}{\sqrt{1 - x^2}} \) is \( \frac{x}{(1 - x^2)^{3/2}} \).
- The derivative of \( \frac{1}{1 + x^2} \) is \( -\frac{2x}{(1 + x^2)^2} \).

Therefore, the new numerator becomes:

\[ \frac{x}{(1 - x^2)^{3/2}} + \frac{2x}{(1 + x^2)^2}. \]

2. Differentiate the denominator \( 3x^2 \) to get \( 6x \).

Now we rewrite our limit as:

\[ \lim_{x \to 0} \frac{\frac{x}{(1 - x^2)^{3/2}} + \frac{2x}{(1 + x^2)^2}}{6x}. \]

### Step 4: Simplify the expression

We can cancel \( x \) from the numerator and denominator (as long as \( x \neq 0 \)):

\[ \lim_{x \to 0} \frac{1}{6} \left( \frac{1}{(1 - x^2)^{3/2}} + \frac{2}{(1 + x^2)^2} \right). \]

### Step 5: Substitute \( x = 0 \) again

Now substituting \( x = 0 \):

\[ \frac{1}{6} \left( \frac{1}{(1 - 0^2)^{3/2}} + \frac{2}{(1 + 0^2)^2} \right) = \frac{1}{6} \left( 1 + 2 \right) = \frac{1}{6} \cdot 3 = \frac{1}{2}. \]

### Final Answer

Thus, the limit is

\[ \boxed{\frac{1}{2}}. \]