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. `(1)/(2)`
C. `0`
D. `1`

Answer

To solve the limit problem given by

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

where \( a, b, c \in \mathbb{R} \) and we want this limit to be a finite number, we will follow these steps:

### 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 gives us an indeterminate form \( \frac{0}{0} \). To resolve this, we need the numerator to also approach 0 as \( x \to 0 \).

### Step 2: Set up the condition for the numerator

For the limit to be finite, the numerator must be 0 when \( x \to 0 \):

\[ \sin(0) + a e^0 + b e^0 + c \ln(1 + 0) = 0 \implies a + b = 0 \implies b = -a \]

### Step 3: Apply L'Hôpital's Rule

Since we still have a \( \frac{0}{0} \) form, we apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator:

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

### Step 4: Evaluate the new limit

Substituting \( x = 0 \) again:

\[ \cos(0) + a - b + c = 1 + a - (-a) + c = 1 + 2a + c \]

The denominator becomes \( 3(0)^2 = 0 \), so we again have a \( \frac{0}{0} \) form. Thus, we need \( 1 + 2a + c = 0 \).

### Step 5: Set up the second condition

From the previous step, we have:

\[ 1 + 2a + c = 0 \implies c = -1 - 2a \]

### Step 6: Differentiate again

We apply L'Hôpital's Rule again:

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

### Step 7: Evaluate the new limit again

Substituting \( x = 0 \):

\[ -\sin(0) + a + b - c = 0 + a - a - (-1 - 2a) = 1 + 2a \]

The denominator is still approaching 0, so we need \( 1 + 2a = 0 \) to ensure the limit is finite.

### Step 8: Solve for \( a, b, c \)

From \( 1 + 2a = 0 \):

\[ 2a = -1 \implies a = -\frac{1}{2} \]

Then substituting back:

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

### Final Values

Thus, we have:

- \( a = -\frac{1}{2} \)
- \( b = \frac{1}{2} \)
- \( c = 0 \)

### Step 9: Final limit evaluation

Now substituting these values back into our limit:

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

Evaluating this limit gives us a finite value, which can be calculated as:

\[ \lim_{x \to 0} f(x) = -\frac{1}{3} \]

### Conclusion

Thus, the final answer is:

\[ \lim_{x \to 0} f(x) = -\frac{1}{3} \]