If f(x)=`(3x^3-7x^2+2)/(4x^2-3x-1)` what does f(x) approach as x gets infinitely larger?

Options

A. 0
B. `3/4`
C. `1`
D. `oo`

Explanation

To find the limit of the function \( f(x) = \frac{3x^3 - 7x^2 + 2}{4x^2 - 3x - 1} \) as \( x \) approaches infinity, we can follow these steps:

### Step 1: Identify the highest power of \( x \)
In the numerator, the highest power of \( x \) is \( x^3 \) (from \( 3x^3 \)), and in the denominator, the highest power of \( x \) is \( x^2 \) (from \( 4x^2 \)).

### Step 2: Divide both the numerator and the denominator by the highest power of \( x \) in the denominator
We will divide both the numerator and the denominator by \( x^3 \) (the highest power in the numerator):

\[ f(x) = \frac{3x^3 - 7x^2 + 2}{4x^2 - 3x - 1} = \frac{3 - \frac{7}{x} + \frac{2}{x^3}}{\frac{4}{x} - \frac{3}{x^2} - \frac{1}{x^3}} \]

### Step 3: Take the limit as \( x \) approaches infinity
Now, we will evaluate the limit as \( x \) approaches infinity:

\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{3 - \frac{7}{x} + \frac{2}{x^3}}{\frac{4}{x} - \frac{3}{x^2} - \frac{1}{x^3}} \]

### Step 4: Evaluate the limit of each term
As \( x \) approaches infinity, the terms \( \frac{7}{x} \), \( \frac{2}{x^3} \), \( \frac{4}{x} \), \( \frac{3}{x^2} \), and \( \frac{1}{x^3} \) all approach 0:

\[ \lim_{x \to \infty} f(x) = \frac{3 - 0 + 0}{0 - 0 - 0} = \frac{3}{0} \]

### Step 5: Determine the behavior of the function
Since we have \( \frac{3}{0} \), this indicates that the function approaches infinity as \( x \) approaches infinity.

### Conclusion
Thus, we conclude that:

\[ \lim_{x \to \infty} f(x) = \infty \]

The correct answer is **infinity**.

---