If `lim_(xrarroo) ((x+c)/(x-c))^(x)=4` then the value of `e^(c)` is

Options

A. `1//4`
B. `1//2`
C. 1
D. 2

Explanation

To solve the limit problem given by the expression \( \lim_{x \to \infty} \left( \frac{x+c}{x-c} \right)^x = 4 \), we will follow these steps:

### Step 1: Rewrite the Limit Expression
We start by rewriting the limit expression:
\[ L = \lim_{x \to \infty} \left( \frac{x+c}{x-c} \right)^x \]

### Step 2: Factor Out \( x \)
Next, we factor \( x \) out of the numerator and denominator:
\[ L = \lim_{x \to \infty} \left( \frac{x(1+\frac{c}{x})}{x(1-\frac{c}{x})} \right)^x = \lim_{x \to \infty} \left( \frac{1+\frac{c}{x}}{1-\frac{c}{x}} \right)^x \]

### Step 3: Identify the Indeterminate Form
As \( x \to \infty \), both \( \frac{c}{x} \) terms approach 0, leading to the form \( \frac{1+0}{1-0} = 1 \). Thus, we have:
\[ L = \lim_{x \to \infty} \left( 1 \right)^x \]
This is an indeterminate form \( 1^\infty \).

### Step 4: Apply the Exponential Limit Formula
To resolve the indeterminate form, we use the formula:
\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} (f(x)-1)g(x)} \]
Here, let \( f(x) = \frac{1+\frac{c}{x}}{1-\frac{c}{x}} \) and \( g(x) = x \).

### Step 5: Simplify \( f(x) - 1 \)
We calculate \( f(x) - 1 \):
\[ f(x) - 1 = \frac{1+\frac{c}{x}}{1-\frac{c}{x}} - 1 = \frac{(1+\frac{c}{x}) - (1-\frac{c}{x})}{1-\frac{c}{x}} = \frac{\frac{2c}{x}}{1-\frac{c}{x}} = \frac{2c}{x(1-\frac{c}{x})} \]

### Step 6: Multiply by \( g(x) \)
Now, we multiply by \( g(x) = x \):
\[ \lim_{x \to \infty} (f(x) - 1)g(x) = \lim_{x \to \infty} \frac{2c}{1-\frac{c}{x}} = 2c \]

### Step 7: Substitute Back into the Exponential Limit
Thus, we have:
\[ L = e^{\lim_{x \to \infty} (f(x) - 1)g(x)} = e^{2c} \]
Given that \( L = 4 \), we set:
\[ e^{2c} = 4 \]

### Step 8: Solve for \( c \)
Taking the natural logarithm of both sides:
\[ 2c = \ln(4) \]
\[ c = \frac{1}{2} \ln(4) \]

### Step 9: Find \( e^c \)
To find \( e^c \):
\[ e^c = e^{\frac{1}{2} \ln(4)} = \sqrt{4} = 2 \]

### Final Answer
Thus, the value of \( e^c \) is:
\[ \boxed{2} \]