If `L=lim_(xto0) (sinx+ae^(x)+be^(-x)+clog_(e)(1+x))/(x^(3))` exists finitely, then
The value of L is

Explanation

To solve the limit problem given, we start with the expression:

\[ L = \lim_{x \to 0} \frac{\sin x + ae^x + be^{-x} + c \log_e(1+x)}{x^3} \]

### Step 1: Expand each term using Taylor series

1. **Sine function**:
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]

2. **Exponential functions**:
\[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4) \]
\[ e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + O(x^4) \]

3. **Logarithmic function**:
\[ \log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4) \]

### Step 2: Substitute the expansions into the limit expression

Substituting these expansions into the limit gives:

\[ L = \lim_{x \to 0} \frac{\left(x - \frac{x^3}{6}\right) + a\left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right) + b\left(1 - x + \frac{x^2}{2} - \frac{x^3}{6}\right) + c\left(x - \frac{x^2}{2} + \frac{x^3}{3}\right)}{x^3} \]

### Step 3: Combine like terms

Combining the constant, linear, quadratic, and cubic terms:

- **Constant term**:
\[ a + b \]

- **Linear term**:
\[ 1 + a - b + c \]

- **Quadratic term**:
\[ \frac{a}{2} + \frac{b}{2} - \frac{c}{2} \]

- **Cubic term**:
\[ -\frac{1}{6} + \frac{a}{6} - \frac{b}{6} + \frac{c}{3} \]

Thus, the expression becomes:

\[ L = \lim_{x \to 0} \frac{(a + b) + (1 + a - b + c)x + \left(\frac{a}{2} + \frac{b}{2} - \frac{c}{2}\right)x^2 + \left(-\frac{1}{6} + \frac{a}{6} - \frac{b}{6} + \frac{c}{3}\right)x^3}{x^3} \]

### Step 4: Set coefficients of lower powers of x to zero

For the limit \(L\) to exist finitely, the coefficients of \(x^0\), \(x^1\), and \(x^2\) must be zero:

1. **Coefficient of \(x^0\)**:
\[ a + b = 0 \quad (1) \]

2. **Coefficient of \(x^1\)**:
\[ 1 + a - b + c = 0 \quad (2) \]

3. **Coefficient of \(x^2\)**:
\[ \frac{a}{2} + \frac{b}{2} - \frac{c}{2} = 0 \quad (3) \]

### Step 5: Solve the system of equations

From equation (1), we have:
\[ b = -a \]

Substituting \(b = -a\) into equation (2):
\[ 1 + a - (-a) + c = 0 \implies 1 + 2a + c = 0 \quad (4) \]

Substituting \(b = -a\) into equation (3):
\[ \frac{a}{2} - \frac{a}{2} - \frac{c}{2} = 0 \implies -\frac{c}{2} = 0 \implies c = 0 \quad (5) \]

Substituting \(c = 0\) into equation (4):
\[ 1 + 2a = 0 \implies a = -\frac{1}{2} \]

Then substituting \(a\) back into equation (1):
\[ b = -(-\frac{1}{2}) = \frac{1}{2} \]

### Step 6: Calculate the limit \(L\)

Now substituting \(a\), \(b\), and \(c\) back into the cubic term:

\[ L = \frac{-\frac{1}{6} - \frac{(-\frac{1}{2})}{6} - \frac{\frac{1}{2}}{6} + \frac{0}{3}}{1} = \frac{-\frac{1}{6} + \frac{1}{12} - \frac{1}{12}}{1} = \frac{-\frac{1}{6}}{1} = -\frac{1}{3} \]

### Final Result:
\[ L = -\frac{1}{3} \]