If `L=lim_(xto0)(sinx+ae^(x)+be^(-x)+clog_(e)(1+x))/(x^(3))` exists finitely, then
Equation `ax^(2)+bx+c=0` has

Explanation

To solve the problem step by step, we will evaluate the limit and derive the conditions for the coefficients \( a \), \( b \), and \( c \) such that the limit exists finitely.

### Step 1: Write the limit expression
We start with the limit expression:
\[ L = \lim_{x \to 0} \frac{\sin x + a e^x + b e^{-x} + c \ln(1+x)}{x^3} \]

### Step 2: Expand the functions using Taylor series
We will expand each function in the numerator using Taylor series around \( x = 0 \):

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

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

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

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

### Step 3: Substitute the expansions into the limit
Substituting these expansions into the limit gives:
\[ \sin x + a e^x + b e^{-x} + c \ln(1+x) = \left( x - \frac{x^3}{6} \right) + a \left( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \right) + b \left( 1 - x + \frac{x^2}{2} - \frac{x^3}{6} \right) + c \left( x - \frac{x^2}{2} + \frac{x^3}{3} \right) \]

Combining the terms:
\[ = (a + b + c)x + \left( a - b - \frac{c}{2} \right)x^2 + \left( -\frac{1}{6} + \frac{a}{6} - \frac{b}{6} + \frac{c}{3} \right)x^3 + O(x^4) \]

### Step 4: Formulate the limit expression
Now, we can write:
\[ L = \lim_{x \to 0} \frac{(a + b + c)x + \left( a - b - \frac{c}{2} \right)x^2 + \left( -\frac{1}{6} + \frac{a}{6} - \frac{b}{6} + \frac{c}{3} \right)x^3 + O(x^4)}{x^3} \]

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

The coefficient of \( x^3 \) will determine the limit:
3. \( -\frac{1}{6} + \frac{a}{6} - \frac{b}{6} + \frac{c}{3} = 0 \)

### Step 6: Solve the equations
From the equations:
1. \( a + b + c = 0 \)
2. \( a - b - \frac{c}{2} = 0 \)
3. \( -\frac{1}{6} + \frac{a}{6} - \frac{b}{6} + \frac{c}{3} = 0 \)

We can solve these equations step by step.

From equation (1), we can express \( c \):
\[ c = -a - b \]

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

Substituting \( b = 3a \) into equation (1):
\[ a + 3a + c = 0 \implies 4a + c = 0 \implies c = -4a \]

Now substituting \( b \) and \( c \) into equation (3):
\[ -\frac{1}{6} + \frac{a}{6} - \frac{3a}{6} + \frac{-4a}{3} = 0 \]
Multiply through by 6 to eliminate the fractions:
\[ -1 + a - 3a - 8a = 0 \implies -1 - 10a = 0 \implies a = -\frac{1}{10} \]

Now substituting back to find \( b \) and \( c \):
\[ b = 3(-\frac{1}{10}) = -\frac{3}{10}, \quad c = -4(-\frac{1}{10}) = \frac{4}{10} = \frac{2}{5} \]

### Step 7: Form the quadratic equation
The quadratic equation becomes:
\[ -\frac{1}{10}x^2 - \frac{3}{10}x + \frac{2}{5} = 0 \]
Multiplying through by -10 to eliminate the fractions:
\[ x^2 + 3x - 4 = 0 \]

### Step 8: Factor the quadratic equation
Factoring gives:
\[ (x + 4)(x - 1) = 0 \]
Thus, the roots are:
\[ x = -4 \quad \text{and} \quad x = 1 \]

### Conclusion
The quadratic equation \( ax^2 + bx + c = 0 \) has two real roots.