If `f(x)=((x^2+5x+3)/(x^2+x+2))^x` then `lim_(xrarroo)f(x)` is equal to

Options

A. `e^4`
B. `e^3`
C. `e^2`
D. `24`

Answer

To solve the limit \( \lim_{x \to \infty} f(x) \) where \( f(x) = \left( \frac{x^2 + 5x + 3}{x^2 + x + 2} \right)^x \), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \left( \frac{x^2 + 5x + 3}{x^2 + x + 2} \right)^x \]

### Step 2: Factor out the highest power of \( x \)
To simplify the expression, we factor out \( x^2 \) from both the numerator and the denominator:
\[ f(x) = \left( \frac{x^2(1 + \frac{5}{x} + \frac{3}{x^2})}{x^2(1 + \frac{1}{x} + \frac{2}{x^2})} \right)^x = \left( \frac{1 + \frac{5}{x} + \frac{3}{x^2}}{1 + \frac{1}{x} + \frac{2}{x^2}} \right)^x \]

### Step 3: Evaluate the limit as \( x \to \infty \)
As \( x \) approaches infinity, the terms \( \frac{5}{x} \), \( \frac{3}{x^2} \), \( \frac{1}{x} \), and \( \frac{2}{x^2} \) approach 0. Therefore, we have:
\[ \lim_{x \to \infty} f(x) = \left( \frac{1 + 0 + 0}{1 + 0 + 0} \right)^x = 1^x \]
This is an indeterminate form of \( 1^\infty \).

### Step 4: Use the exponential limit form
To resolve the \( 1^\infty \) form, we can use the formula:
\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} g(x) \cdot (f(x) - 1)} \]
In our case, \( g(x) = x \) and \( f(x) = \frac{1 + \frac{5}{x} + \frac{3}{x^2}}{1 + \frac{1}{x} + \frac{2}{x^2}} \).

### Step 5: Calculate \( f(x) - 1 \)
We need to find:
\[ f(x) - 1 = \frac{1 + \frac{5}{x} + \frac{3}{x^2}}{1 + \frac{1}{x} + \frac{2}{x^2}} - 1 = \frac{(1 + \frac{5}{x} + \frac{3}{x^2}) - (1 + \frac{1}{x} + \frac{2}{x^2})}{1 + \frac{1}{x} + \frac{2}{x^2}} \]
Simplifying the numerator:
\[ = \frac{\frac{5}{x} + \frac{3}{x^2} - \frac{1}{x} - \frac{2}{x^2}}{1 + \frac{1}{x} + \frac{2}{x^2}} = \frac{\frac{4}{x} + \frac{1}{x^2}}{1 + \frac{1}{x} + \frac{2}{x^2}} \]

### Step 6: Multiply by \( g(x) \)
Now, we multiply by \( g(x) = x \):
\[ x \cdot (f(x) - 1) = x \cdot \frac{\frac{4}{x} + \frac{1}{x^2}}{1 + \frac{1}{x} + \frac{2}{x^2}} = \frac{4 + \frac{1}{x}}{1 + \frac{1}{x} + \frac{2}{x^2}} \]

### Step 7: Take the limit as \( x \to \infty \)
Taking the limit:
\[ \lim_{x \to \infty} \frac{4 + \frac{1}{x}}{1 + \frac{1}{x} + \frac{2}{x^2}} = \frac{4 + 0}{1 + 0 + 0} = 4 \]

### Step 8: Final result
Thus, we have:
\[ \lim_{x \to \infty} f(x) = e^{4} \]

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