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 will 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: Simplify the fraction inside the limit
As \( x \) approaches infinity, we can 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 of the fraction
Now, we take the limit of the fraction as \( x \) approaches infinity:
\[ \lim_{x \to \infty} \frac{1 + \frac{5}{x} + \frac{3}{x^2}}{1 + \frac{1}{x} + \frac{2}{x^2}} = \frac{1 + 0 + 0}{1 + 0 + 0} = 1 \]

### Step 4: Identify the indeterminate form
Since we have \( f(x) \) approaching \( 1^x \) as \( x \to \infty \), this is an indeterminate form \( 1^\infty \).

### Step 5: Apply the exponential limit
To resolve the indeterminate form, we use the property:
\[ \lim_{x \to a} f(x)^g(x) = e^{\lim_{x \to a} g(x) \cdot (f(x) - 1)} \]
where \( g(x) = x \) and \( f(x) = \frac{x^2 + 5x + 3}{x^2 + x + 2} \).

### Step 6: Find \( f(x) - 1 \)
We need to find:
\[ f(x) - 1 = \frac{x^2 + 5x + 3}{x^2 + x + 2} - 1 = \frac{x^2 + 5x + 3 - (x^2 + x + 2)}{x^2 + x + 2} = \frac{4x + 1}{x^2 + x + 2} \]

### Step 7: Multiply by \( g(x) \)
Now we multiply by \( g(x) = x \):
\[ g(x) \cdot (f(x) - 1) = x \cdot \frac{4x + 1}{x^2 + x + 2} = \frac{4x^2 + x}{x^2 + x + 2} \]

### Step 8: Find the limit of \( g(x) \cdot (f(x) - 1) \)
Now, we take the limit as \( x \to \infty \):
\[ \lim_{x \to \infty} \frac{4x^2 + x}{x^2 + x + 2} = \lim_{x \to \infty} \frac{4 + \frac{1}{x}}{1 + \frac{1}{x} + \frac{2}{x^2}} = \frac{4 + 0}{1 + 0 + 0} = 4 \]

### Step 9: Apply the exponential function
Finally, we can substitute back into the exponential limit:
\[ \lim_{x \to \infty} f(x) = e^{\lim_{x \to \infty} g(x) \cdot (f(x) - 1)} = e^4 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to \infty} f(x) = e^4 \]