Let `lim_(x to 0) ` f(x) be a finite number, where
`f(x) = ("sin" x + ae^(x) + be^(-x) + cln (1 + x))/(x^(3))` a,b,e in R`

Options

A. `-(1)/(3)`
B. 0
C. `(1)/(2)`
D. 1

Answer

To solve the problem, we need to find the conditions under which the limit

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

is finite.

### Step 1: Evaluate the limit at \(x = 0\)

First, we substitute \(x = 0\) into the function:

\[ f(0) = \frac{\sin(0) + a e^0 + b e^0 + c \ln(1+0)}{0^3} = \frac{0 + a + b + 0}{0} \]

This results in the indeterminate form \(\frac{0}{0}\). To resolve this, we 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:

- **Numerator:**
\[ \frac{d}{dx}(\sin x + ae^x + be^{-x} + c\ln(1+x)) = \cos x + ae^x - be^{-x} + \frac{c}{1+x} \]

- **Denominator:**
\[ \frac{d}{dx}(x^3) = 3x^2 \]

Now we have:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\cos x + ae^x - be^{-x} + \frac{c}{1+x}}{3x^2} \]

### Step 3: Evaluate the limit again at \(x = 0\)

Substituting \(x = 0\) again gives:

\[ \frac{\cos(0) + a - b + c}{0} = \frac{1 + a - b + c}{0} \]

This is still an indeterminate form \(\frac{0}{0}\). Therefore, we apply L'Hôpital's Rule again.

### Step 4: Differentiate again

Differentiate the numerator and denominator again:

- **Numerator:**
\[ \frac{d}{dx}(\cos x + ae^x - be^{-x} + \frac{c}{1+x}) = -\sin x + ae^x + be^{-x} - \frac{c}{(1+x)^2} \]

- **Denominator:**
\[ \frac{d}{dx}(3x^2) = 6x \]

Now we have:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{-\sin x + ae^x + be^{-x} - \frac{c}{(1+x)^2}}{6x} \]

### Step 5: Evaluate the limit again at \(x = 0\)

Substituting \(x = 0\) gives:

\[ \frac{0 + a + b - c}{0} = \frac{a + b - c}{0} \]

This is still indeterminate. Thus, we need to ensure that the numerator approaches 0 as \(x\) approaches 0.

### Step 6: Set conditions for the limit to exist

From our earlier steps, we derived two conditions:

1. \(1 + a - b + c = 0\)
2. \(a + b - c = 0\)

### Step 7: Solve the equations

From the first equation, we can express \(c\):

\[ c = -1 - a + b \]

Substituting \(c\) into the second equation:

\[ a + b - (-1 - a + b) = 0 \implies a + b + 1 + a - b = 0 \implies 2a + 1 = 0 \implies a = -\frac{1}{2} \]

Substituting \(a\) back to find \(b\):

\[ b = \frac{1}{2} - c \]

Substituting \(a\) into the first equation gives:

\[ 1 - \frac{1}{2} - b + c = 0 \implies \frac{1}{2} - b + c = 0 \]

### Final Step: Conclusion

After solving these equations, we find that:

\[ c = 0, \quad a = -\frac{1}{2}, \quad b = \frac{1}{2} \]

Thus, the final answer is:

\[ \boxed{-\frac{1}{3}} \]