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. 2

Answer

To solve the limit problem given by

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

where \( a, b, c \in \mathbb{R} \) and the limit is finite, we will follow these steps:

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

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

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

The denominator becomes:

\[ 0^3 = 0 \]

Since both the numerator and denominator approach 0, we have an indeterminate form \( \frac{0}{0} \). For the limit to be finite, the numerator must also approach 0 as \( x \to 0 \). Therefore, we need:

\[ a + b = 0 \quad \text{(Equation 1)} \]

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule. We differentiate the numerator and denominator:

- The derivative of the numerator:
\[ \frac{d}{dx}(\sin x + ae^x + be^{-x} + c \ln(1+x)) = \cos x + ae^x - be^{-x} + \frac{c}{1+x} \]

- The derivative of the denominator:
\[ \frac{d}{dx}(x^3) = 3x^2 \]

Now we have:

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

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

Substituting \( x = 0 \):

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

The denominator becomes:

\[ 3(0)^2 = 0 \]

Again, we have an indeterminate form \( \frac{0}{0} \). For the limit to be finite, we require:

\[ 1 + a - b + c = 0 \quad \text{(Equation 2)} \]

### Step 4: Substitute \( b \) from Equation 1 into Equation 2

From Equation 1, we have \( b = -a \). Substituting this into Equation 2:

\[ 1 + a - (-a) + c = 0 \implies 1 + 2a + c = 0 \quad \text{(Equation 3)} \]

### Step 5: Apply L'Hôpital's Rule again

Differentiate the numerator and denominator again:

- The derivative of the numerator:
\[ -\sin x + ae^x + be^{-x} - \frac{c}{(1+x)^2} \]

- The derivative of the denominator:
\[ \frac{d}{dx}(3x^2) = 6x \]

Now we have:

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

### Step 6: Evaluate the new limit at \( x = 0 \)

Substituting \( x = 0 \):

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

The denominator becomes:

\[ 6(0) = 0 \]

For the limit to be finite, we need:

\[ -c = 0 \implies c = 0 \quad \text{(Equation 4)} \]

### Step 7: Substitute \( c = 0 \) back into Equation 3

Substituting \( c = 0 \) into Equation 3:

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

### Step 8: Find \( b \)

Using Equation 1:

\[ b = -a = -\left(-\frac{1}{2}\right) = \frac{1}{2} \]

### Final Values

We have determined:

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

### Conclusion

Thus, the final answer for the limit is:

\[ \lim_{x \to 0} \frac{\sin x - \frac{1}{2} e^x + \frac{1}{2} e^{-x}}{x^3} = -\frac{1}{3} \]