If `f(x) = ((x^(2) + 5x + 3)/(x^(2) + x+2))^(x)`, then `lim_(x to infty) f(x)` is:

Options

A. `e^(4)`
B. `e^(3)`
C. `e^(2)`
D. `2^(4)`

Answer

To find the limit of the function \( f(x) = \left( \frac{x^2 + 5x + 3}{x^2 + x + 2} \right)^x \) as \( x \) approaches infinity, we will follow these steps:

### Step-by-step Solution:

1. **Rewrite the Function**:
\[ f(x) = \left( \frac{x^2 + 5x + 3}{x^2 + x + 2} \right)^x \]

2. **Divide Numerator and Denominator by \( x^2 \)**:
\[ f(x) = \left( \frac{1 + \frac{5}{x} + \frac{3}{x^2}}{1 + \frac{1}{x} + \frac{2}{x^2}} \right)^x \]

3. **Take the Limit Inside the Exponent**:
As \( x \to \infty \), the terms \( \frac{5}{x} \), \( \frac{3}{x^2} \), \( \frac{1}{x} \), and \( \frac{2}{x^2} \) approach 0. Thus, we have:
\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left( \frac{1 + 0 + 0}{1 + 0 + 0} \right)^x = 1^x \]

4. **Indeterminate Form**:
Since \( 1^x \) is an indeterminate form, we can use the logarithm to evaluate the limit:
\[ \ln(f(x)) = x \cdot \ln\left( \frac{x^2 + 5x + 3}{x^2 + x + 2} \right) \]

5. **Evaluate the Logarithm**:
We can rewrite the logarithm:
\[ \ln\left( \frac{x^2 + 5x + 3}{x^2 + x + 2} \right) = \ln(x^2 + 5x + 3) - \ln(x^2 + x + 2) \]

6. **Apply L'Hôpital's Rule**:
As \( x \to \infty \), both \( \ln(x^2 + 5x + 3) \) and \( \ln(x^2 + x + 2) \) approach infinity. We can apply L'Hôpital's Rule:
\[ \lim_{x \to \infty} \left( \ln(x^2 + 5x + 3) - \ln(x^2 + x + 2) \right) \]

7. **Differentiate the Numerator and Denominator**:
Differentiate both functions:
\[ \frac{d}{dx}(\ln(x^2 + 5x + 3)) = \frac{2x + 5}{x^2 + 5x + 3}, \quad \frac{d}{dx}(\ln(x^2 + x + 2)) = \frac{2x + 1}{x^2 + x + 2} \]

8. **Take the Limit**:
Now we take the limit:
\[ \lim_{x \to \infty} \left( \frac{2x + 5}{x^2 + 5x + 3} - \frac{2x + 1}{x^2 + x + 2} \right) \]
This simplifies to:
\[ \lim_{x \to \infty} \left( \frac{(2x + 5)(x^2 + x + 2) - (2x + 1)(x^2 + 5x + 3)}{(x^2 + 5x + 3)(x^2 + x + 2)} \right) \]

9. **Final Limit Calculation**:
After simplifying, we find that the limit approaches \( 4 \). Thus:
\[ \lim_{x \to \infty} \ln(f(x)) = 4 \implies \lim_{x \to \infty} f(x) = e^4 \]

### Conclusion:
The limit is:
\[ \lim_{x \to \infty} f(x) = e^4 \]