`lim_(x to infty)((x+3)/(x-1))^(x+3)` =

Options

A. `e^(2)`
B. `e^(4)`
C. `e^(5)`
D. `e^(6)`

Answer

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

### Step 1: Identify the Form
We start by recognizing that as \(x\) approaches infinity, the expression \(\frac{x+3}{x-1}\) approaches 1. Thus, we have an indeterminate form of type \(1^{\infty}\).

### Step 2: Rewrite the Limit
To handle the indeterminate form, we can rewrite the limit using the exponential function:
\[ L = \lim_{x \to \infty} \left(\frac{x+3}{x-1}\right)^{x+3} = e^{\lim_{x \to \infty} (x+3) \ln\left(\frac{x+3}{x-1}\right)} \]

### Step 3: Simplify the Logarithm
Next, we simplify the logarithm:
\[ \ln\left(\frac{x+3}{x-1}\right) = \ln(x+3) - \ln(x-1) \]
Using the properties of logarithms, we can express this as:
\[ \ln\left(\frac{x+3}{x-1}\right) = \ln\left(1 + \frac{4}{x-1}\right) \]
For large \(x\), we can use the approximation \(\ln(1 + u) \approx u\) when \(u\) is small:
\[ \ln\left(1 + \frac{4}{x-1}\right) \approx \frac{4}{x-1} \]

### Step 4: Substitute Back into the Limit
Now substituting back, we have:
\[ \lim_{x \to \infty} (x+3) \ln\left(\frac{x+3}{x-1}\right) \approx \lim_{x \to \infty} (x+3) \cdot \frac{4}{x-1} \]

### Step 5: Simplify the Expression
This simplifies to:
\[ \lim_{x \to \infty} \frac{4(x+3)}{x-1} = \lim_{x \to \infty} \frac{4x + 12}{x - 1} \]
Dividing the numerator and the denominator by \(x\):
\[ = \lim_{x \to \infty} \frac{4 + \frac{12}{x}}{1 - \frac{1}{x}} = \frac{4 + 0}{1 - 0} = 4 \]

### Step 6: Final Result
Now substituting this back into our expression for \(L\):
\[ L = e^4 \]

Thus, the final answer is:
\[ \lim_{x \to \infty} \left(\frac{x+3}{x-1}\right)^{x+3} = e^4 \]