`lim_(xrarr oo) (1-(4)/(x-1))^(3x-1)` is equal to

Options

A. `e^12`
B. `e^-12`
C. `e^4`
D. `e^3`

Explanation

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

### Step 1: Identify the form of the limit
As \( x \) approaches infinity, the expression \( 1 - \frac{4}{x - 1} \) approaches \( 1 \). Therefore, we have the form \( 1^{\infty} \), which is an indeterminate form.

**Hint**: Recognize that \( 1^{\infty} \) is an indeterminate form that can be transformed using logarithms.

### Step 2: Rewrite the limit using the exponential function
We can rewrite the limit in the form of \( e \) by using the property:
\[ a^b = e^{b \ln(a)} \]
Thus, we can express our limit as:
\[ \lim_{x \to \infty} \left( 1 - \frac{4}{x - 1} \right)^{3x - 1} = e^{\lim_{x \to \infty} (3x - 1) \ln\left(1 - \frac{4}{x - 1}\right)} \]

**Hint**: Use the property of logarithms to simplify the expression.

### Step 3: Simplify the logarithm
Now, we need to find \( \ln\left(1 - \frac{4}{x - 1}\right) \). For small values of \( y \), we can use the approximation \( \ln(1 - y) \approx -y \):
\[ \ln\left(1 - \frac{4}{x - 1}\right) \approx -\frac{4}{x - 1} \]
Thus, we have:
\[ \lim_{x \to \infty} (3x - 1) \ln\left(1 - \frac{4}{x - 1}\right) \approx \lim_{x \to \infty} (3x - 1) \left(-\frac{4}{x - 1}\right) \]

**Hint**: Use the approximation for logarithms when \( x \) is large.

### Step 4: Simplify the limit expression
Now we simplify the limit:
\[ \lim_{x \to \infty} (3x - 1) \left(-\frac{4}{x - 1}\right) = -4 \lim_{x \to \infty} \frac{3x - 1}{x - 1} \]
We can simplify \( \frac{3x - 1}{x - 1} \) as follows:
\[ \frac{3x - 1}{x - 1} = \frac{3 - \frac{1}{x}}{1 - \frac{1}{x}} \to 3 \quad \text{as } x \to \infty \]
Thus, we have:
\[ -4 \cdot 3 = -12 \]

**Hint**: Factor out the leading terms to find the limit as \( x \) approaches infinity.

### Step 5: Final result
Putting it all together, we find:
\[ \lim_{x \to \infty} \left( 1 - \frac{4}{x - 1} \right)^{3x - 1} = e^{-12} \]

Therefore, the final answer is:
\[ \boxed{e^{-12}} \]