If `L=lim_(xto0) (1)/(x^(3))((1)/(sqrt(1+x))-(1+ax)/(1+bx))` exists,then

Options

A. `cosLltcosR`
B. `tan(2L)ltanlt2R`
C. `sin LgtsinR`
D. `None of these

Explanation

To solve the limit problem \( L = \lim_{x \to 0} \frac{1}{x^3} \left( \frac{1}{\sqrt{1+x}} - \frac{1+ax}{1+bx} \right) \) and determine the conditions under which it exists, we can follow these steps:

### Step 1: Simplify the expression inside the limit
We start with the expression:
\[ L = \lim_{x \to 0} \frac{1}{x^3} \left( \frac{1}{\sqrt{1+x}} - \frac{1+ax}{1+bx} \right) \]

### Step 2: Expand the terms using Taylor series
We can use the Taylor series expansion for small values of \( x \):
- For \( \sqrt{1+x} \):
\[ \sqrt{1+x} \approx 1 + \frac{x}{2} - \frac{x^2}{8} + O(x^3) \]
Thus,
\[ \frac{1}{\sqrt{1+x}} \approx 1 - \frac{x}{2} + \frac{x^2}{8} + O(x^3) \]

- For \( \frac{1+ax}{1+bx} \):
Using the expansion:
\[ \frac{1+ax}{1+bx} \approx (1 + ax)(1 - bx + b^2x^2 - \ldots) \approx 1 + (a-b)x + (b^2 - ab)x^2 + O(x^3) \]

### Step 3: Substitute the expansions back into the limit
Now substituting these expansions into our limit:
\[ L = \lim_{x \to 0} \frac{1}{x^3} \left( \left( 1 - \frac{x}{2} + \frac{x^2}{8} \right) - \left( 1 + (a-b)x + (b^2 - ab)x^2 \right) \right) \]

### Step 4: Combine like terms
Combining the terms:
\[ L = \lim_{x \to 0} \frac{1}{x^3} \left( -\left(\frac{1}{2} + (a-b)\right)x + \left(\frac{1}{8} - (b^2 - ab)\right)x^2 \right) \]

### Step 5: Analyze the limit
For the limit \( L \) to exist, the coefficients of \( x \) and \( x^2 \) must be zero:
1. Coefficient of \( x \):
\[ -\left(\frac{1}{2} + (a-b)\right) = 0 \implies a - b = -\frac{1}{2} \]

2. Coefficient of \( x^2 \):
\[ \frac{1}{8} - (b^2 - ab) = 0 \implies b^2 - ab = \frac{1}{8} \]

### Step 6: Solve the equations
Now we have two equations:
1. \( a - b = -\frac{1}{2} \)
2. \( b^2 - ab = \frac{1}{8} \)

From the first equation, we can express \( a \) in terms of \( b \):
\[ a = b - \frac{1}{2} \]

Substituting this into the second equation:
\[ b^2 - b(b - \frac{1}{2}) = \frac{1}{8} \]
This simplifies to:
\[ b^2 - b^2 + \frac{b}{2} = \frac{1}{8} \implies \frac{b}{2} = \frac{1}{8} \implies b = \frac{1}{4} \]

Now substituting \( b \) back to find \( a \):
\[ a = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \]

### Conclusion
Thus, the values of \( a \) and \( b \) for which the limit exists are:
\[ a = -\frac{1}{4}, \quad b = \frac{1}{4} \]