`lim_(xrarr oo) ((3x^2+2x+1)/(x^2+x+2))^((6x+1)/(3x+1))` , is equal to

Options

A. 3
B. 6
C. 9
D. none of these

Answer

To solve the limit \( \lim_{x \to \infty} \left( \frac{3x^2 + 2x + 1}{x^2 + x + 2} \right)^{\frac{6x + 1}{3x + 1}} \), we will follow these steps:

### Step 1: Simplify the expression inside the limit

We start with the expression:

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

To simplify this, we can factor out \( x^2 \) from both the numerator and the denominator:

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

This simplifies to:

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

### Step 2: Take the limit as \( x \to \infty \)

As \( x \) approaches infinity, the terms \( \frac{2}{x} \), \( \frac{1}{x^2} \), \( \frac{1}{x} \), and \( \frac{2}{x^2} \) all approach 0. Thus, we have:

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

### Step 3: Simplify the exponent

Now, we need to simplify the exponent:

\[ \frac{6x + 1}{3x + 1} \]

Factoring out \( x \) from both the numerator and the denominator gives:

\[ = \frac{x(6 + \frac{1}{x})}{x(3 + \frac{1}{x})} = \frac{6 + \frac{1}{x}}{3 + \frac{1}{x}} \]

Taking the limit as \( x \to \infty \):

\[ \lim_{x \to \infty} \frac{6 + \frac{1}{x}}{3 + \frac{1}{x}} = \frac{6 + 0}{3 + 0} = \frac{6}{3} = 2 \]

### Step 4: Combine the results

Now we can combine the results from Steps 2 and 3:

\[ \lim_{x \to \infty} \left( \frac{3x^2 + 2x + 1}{x^2 + x + 2} \right)^{\frac{6x + 1}{3x + 1}} = 3^2 = 9 \]

### Final Answer

Thus, the limit is:

\[ \boxed{9} \]