`lim_(x to oo) ((x^(2)+ 5x+3)/(x^(2)+x+3))^(x)` is equal to

Options

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

Answer

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

### Step 1: Identify the Form of the Limit
First, we need to analyze the expression inside the limit as \( x \) approaches infinity. We can rewrite the limit as:

\[ \lim_{x \to \infty} \left( \frac{x^2 + 5x + 3}{x^2 + x + 3} \right)^x \]

### Step 2: Simplify the Fraction
To simplify the fraction, we divide both the numerator and the denominator by \( x^2 \):

\[ \frac{x^2 + 5x + 3}{x^2 + x + 3} = \frac{1 + \frac{5}{x} + \frac{3}{x^2}}{1 + \frac{1}{x} + \frac{3}{x^2}} \]

### Step 3: Evaluate the Limit of the Fraction
Now, as \( x \) approaches infinity, the terms \( \frac{5}{x} \), \( \frac{3}{x^2} \), and \( \frac{1}{x} \) all approach 0. Thus, we have:

\[ \lim_{x \to \infty} \frac{x^2 + 5x + 3}{x^2 + x + 3} = \frac{1 + 0 + 0}{1 + 0 + 0} = 1 \]

### Step 4: Identify the Indeterminate Form
Since we are looking at the limit of the form \( 1^\infty \), we can apply the following transformation:

\[ \lim_{x \to \infty} \left( \frac{x^2 + 5x + 3}{x^2 + x + 3} \right)^x = e^{\lim_{x \to \infty} \left( \frac{x^2 + 5x + 3}{x^2 + x + 3} - 1 \right) x} \]

### Step 5: Simplify the Expression Inside the Exponent
Next, we need to simplify \( \frac{x^2 + 5x + 3}{x^2 + x + 3} - 1 \):

\[ \frac{x^2 + 5x + 3 - (x^2 + x + 3)}{x^2 + x + 3} = \frac{(5x - x)}{x^2 + x + 3} = \frac{4x}{x^2 + x + 3} \]

### Step 6: Multiply by \( x \)
Now, we multiply this by \( x \):

\[ \lim_{x \to \infty} \frac{4x^2}{x^2 + x + 3} \]

### Step 7: Simplify Again
We can divide the numerator and the denominator by \( x^2 \):

\[ \lim_{x \to \infty} \frac{4}{1 + \frac{1}{x} + \frac{3}{x^2}} = \frac{4}{1 + 0 + 0} = 4 \]

### Step 8: Final Limit Calculation
Now we can substitute this back into our exponent:

\[ e^{\lim_{x \to \infty} \left( \frac{x^2 + 5x + 3}{x^2 + x + 3} - 1 \right) x} = e^4 \]

### Conclusion
Thus, the limit is:

\[ \lim_{x \to \infty} \left( \frac{x^2 + 5x + 3}{x^2 + x + 3} \right)^x = e^4 \]