The value of `lim_(x rarr oo) ((3x-4)/(3x+2))` is

Options

A. `e^(-1//3)`
B. `e^(-2//3)`
C. `e^(-1)`
D. `1`

Answer

To find the value of the limit

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

we will follow these steps:

### Step 1: Identify the limit expression
We start with the expression

\[ \frac{3x - 4}{3x + 2} \]

and we need to evaluate the limit as \( x \) approaches infinity.

### Step 2: Factor out \( x \) from the numerator and denominator
To simplify the expression, we can factor \( x \) out of both the numerator and the denominator:

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

### Step 3: Cancel \( x \) from the numerator and denominator
Since \( x \) is common in both the numerator and the denominator, we can cancel it out:

\[ = \frac{3 - \frac{4}{x}}{3 + \frac{2}{x}} \]

### Step 4: Evaluate the limit as \( x \to \infty \)
Now we take the limit of the simplified expression as \( x \) approaches infinity:

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

As \( x \) approaches infinity, the terms \( \frac{4}{x} \) and \( \frac{2}{x} \) approach 0. Therefore, we have:

\[ = \frac{3 - 0}{3 + 0} = \frac{3}{3} = 1. \]

### Conclusion
Thus, the value of the limit is

\[ \boxed{1}. \]

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