`lim_(xrarr0) (sin^-1x-tan^-1x)/(3x^3)=L` then find (6L+1)

Answer

To solve the limit \( L = \lim_{x \to 0} \frac{\sin^{-1} x - \tan^{-1} x}{3x^3} \), we will follow these steps:

### Step 1: Evaluate the limit directly
First, we substitute \( x = 0 \):
\[ \sin^{-1}(0) = 0 \quad \text{and} \quad \tan^{-1}(0) = 0 \]
Thus, we have:
\[ L = \frac{0 - 0}{3 \cdot 0^3} = \frac{0}{0} \]
This is an indeterminate form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:
- The derivative of \( \sin^{-1} x \) is \( \frac{1}{\sqrt{1 - x^2}} \).
- The derivative of \( \tan^{-1} x \) is \( \frac{1}{1 + x^2} \).
- The derivative of \( 3x^3 \) is \( 9x^2 \).

Now we apply L'Hôpital's Rule:
\[ L = \lim_{x \to 0} \frac{\frac{1}{\sqrt{1 - x^2}} - \frac{1}{1 + x^2}}{9x^2} \]

### Step 3: Simplify the expression
To simplify the expression in the numerator, we find a common denominator:
\[ \frac{1}{\sqrt{1 - x^2}} - \frac{1}{1 + x^2} = \frac{(1 + x^2) - \sqrt{1 - x^2}}{\sqrt{1 - x^2}(1 + x^2)} \]
Thus, we have:
\[ L = \lim_{x \to 0} \frac{(1 + x^2) - \sqrt{1 - x^2}}{9x^2 \sqrt{1 - x^2}(1 + x^2)} \]

### Step 4: Substitute \( x = 0 \) again
Substituting \( x = 0 \) gives:
\[ 1 + 0^2 - \sqrt{1 - 0^2} = 1 - 1 = 0 \]
This again results in \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 5: Differentiate again
Differentiating the numerator:
- The derivative of \( (1 + x^2) \) is \( 2x \).
- The derivative of \( \sqrt{1 - x^2} \) using the chain rule is \( -\frac{x}{\sqrt{1 - x^2}} \).

Thus, the derivative of the numerator is:
\[ 2x + \frac{x}{\sqrt{1 - x^2}} \]
The denominator's derivative is:
\[ \frac{d}{dx}(9x^2 \sqrt{1 - x^2}(1 + x^2)) = 9(2x\sqrt{1 - x^2}(1 + x^2) + x^2 \cdot \frac{-x}{\sqrt{1 - x^2}}(1 + x^2) + 9x^2(1 + x^2)) \]

### Step 6: Evaluate the limit
After simplifying, we can substitute \( x = 0 \) again:
\[ L = \lim_{x \to 0} \frac{2 + \frac{1}{1}}{18} = \frac{3}{18} = \frac{1}{6} \]

### Step 7: Calculate \( 6L + 1 \)
Now, we calculate \( 6L + 1 \):
\[ 6L + 1 = 6 \cdot \frac{1}{6} + 1 = 1 + 1 = 2 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{2} \]