`lim_(x rarr infty) (x^(3) int_(-1//x)^(1//x)("In" (1+t^(2)))/(1+e^(t)) dt)` is equal to `

Options

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

Answer

To solve the limit

\[ \lim_{x \to \infty} \left( x^3 \int_{-\frac{1}{x}}^{\frac{1}{x}} \frac{\ln(1+t^2)}{1+e^t} \, dt \right), \]

we will follow these steps:

### Step 1: Analyze the integral

First, we need to analyze the behavior of the integral

\[ \int_{-\frac{1}{x}}^{\frac{1}{x}} \frac{\ln(1+t^2)}{1+e^t} \, dt \]

as \( x \to \infty \). As \( x \) approaches infinity, the limits of integration approach 0.

### Step 2: Approximate the integrand near \( t = 0 \)

Next, we can approximate the integrand for small values of \( t \):

1. For small \( t \), \( \ln(1+t^2) \approx t^2 \) (using the Taylor expansion).
2. Also, \( e^t \approx 1 + t \) for small \( t \).

Thus, the integrand becomes:

\[ \frac{\ln(1+t^2)}{1+e^t} \approx \frac{t^2}{1 + (1 + t)} = \frac{t^2}{2 + t} \approx \frac{t^2}{2} \quad \text{as } t \to 0. \]

### Step 3: Evaluate the integral

Now we evaluate the integral:

\[ \int_{-\frac{1}{x}}^{\frac{1}{x}} \frac{t^2}{2} \, dt = \frac{1}{2} \int_{-\frac{1}{x}}^{\frac{1}{x}} t^2 \, dt. \]

Calculating the integral:

\[ \int t^2 \, dt = \frac{t^3}{3} \quad \text{and evaluating from } -\frac{1}{x} \text{ to } \frac{1}{x}: \]

\[ \int_{-\frac{1}{x}}^{\frac{1}{x}} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{-\frac{1}{x}}^{\frac{1}{x}} = \frac{1}{3} \left( \left(\frac{1}{x}\right)^3 - \left(-\frac{1}{x}\right)^3 \right) = \frac{1}{3} \left( \frac{1}{x^3} + \frac{1}{x^3} \right) = \frac{2}{3x^3}. \]

Thus,

\[ \int_{-\frac{1}{x}}^{\frac{1}{x}} \frac{t^2}{2} \, dt = \frac{1}{2} \cdot \frac{2}{3x^3} = \frac{1}{3x^3}. \]

### Step 4: Substitute back into the limit

Now substituting this back into the limit:

\[ \lim_{x \to \infty} \left( x^3 \cdot \frac{1}{3x^3} \right) = \lim_{x \to \infty} \frac{1}{3} = \frac{1}{3}. \]

### Conclusion

Thus, the limit evaluates to

\[ \lim_{x \to \infty} \left( x^3 \int_{-\frac{1}{x}}^{\frac{1}{x}} \frac{\ln(1+t^2)}{1+e^t} \, dt \right) = \frac{1}{3}. \]

### Final Answer

The final answer is:

\[ \frac{1}{3}. \]