The value of `lim_(xrarroo) ((3x-4)/(3x+2))^(((x+1)/3))` is

Options

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

Answer

To solve the limit \( \lim_{x \to \infty} \left(\frac{3x-4}{3x+2}\right)^{\frac{x+1}{3}} \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the limit:
\[ \lim_{x \to \infty} \left(\frac{3x-4}{3x+2}\right)^{\frac{x+1}{3}} \]

### Step 2: Factor out \( x \) from the numerator and denominator
To simplify the fraction, we factor \( x \) from both the numerator and denominator:
\[ \frac{3x-4}{3x+2} = \frac{x(3 - \frac{4}{x})}{x(3 + \frac{2}{x})} = \frac{3 - \frac{4}{x}}{3 + \frac{2}{x}} \]

### Step 3: Take the limit of the fraction as \( x \to \infty \)
As \( x \) approaches infinity, the terms \( \frac{4}{x} \) and \( \frac{2}{x} \) approach 0:
\[ \lim_{x \to \infty} \frac{3 - \frac{4}{x}}{3 + \frac{2}{x}} = \frac{3 - 0}{3 + 0} = 1 \]

### Step 4: Substitute back into the limit
Now we substitute this back into the limit:
\[ \lim_{x \to \infty} \left(1\right)^{\frac{x+1}{3}} = 1^{\infty} \]
This is an indeterminate form.

### Step 5: Apply the logarithmic limit method
We can use the property of limits for indeterminate forms \( 1^\infty \):
\[ L = \lim_{x \to \infty} \left(\frac{3x-4}{3x+2}\right)^{\frac{x+1}{3}} = e^{\lim_{x \to \infty} \frac{x+1}{3} \left(\frac{3x-4}{3x+2} - 1\right)} \]

### Step 6: Simplify the expression inside the limit
Now we simplify \( \frac{3x-4}{3x+2} - 1 \):
\[ \frac{3x-4}{3x+2} - 1 = \frac{3x-4 - (3x+2)}{3x+2} = \frac{-6}{3x+2} \]

### Step 7: Substitute back into the limit
Now we have:
\[ L = e^{\lim_{x \to \infty} \frac{x+1}{3} \cdot \frac{-6}{3x+2}} \]

### Step 8: Simplify the limit
This can be simplified as follows:
\[ \lim_{x \to \infty} \frac{x+1}{3} \cdot \frac{-6}{3x+2} = \lim_{x \to \infty} \frac{-6(x+1)}{3(3x+2)} = \lim_{x \to \infty} \frac{-6}{9} = -\frac{2}{3} \]

### Step 9: Final result
Thus, we have:
\[ L = e^{-\frac{2}{3}} \]

So, the value of the limit is:
\[ \lim_{x \to \infty} \left(\frac{3x-4}{3x+2}\right)^{\frac{x+1}{3}} = e^{-\frac{2}{3}} \]