The value of `lim_(xrarroo) ((x-1)/(x+1))^(x)`, is

Options

A. 0
B. `e^-1`
C. `e^-2`
D. `e^-3`

Answer

To find the value of the limit \( \lim_{x \to \infty} \left(\frac{x-1}{x+1}\right)^x \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the limit:
\[ \lim_{x \to \infty} \left(\frac{x-1}{x+1}\right)^x \]
We can simplify the fraction inside the limit:
\[ \frac{x-1}{x+1} = \frac{x(1 - \frac{1}{x})}{x(1 + \frac{1}{x})} = \frac{1 - \frac{1}{x}}{1 + \frac{1}{x}} \]

### Step 2: Substitute into the limit
Now, substituting this back into the limit gives us:
\[ \lim_{x \to \infty} \left(\frac{1 - \frac{1}{x}}{1 + \frac{1}{x}}\right)^x \]

### Step 3: Identify the indeterminate form
As \( x \to \infty \), both \( \frac{1 - \frac{1}{x}}{1 + \frac{1}{x}} \) approaches 1. Therefore, we have the form \( 1^\infty \), which is indeterminate.

### Step 4: Use logarithmic transformation
To resolve this indeterminate form, we can take the natural logarithm:
Let \( y = \left(\frac{1 - \frac{1}{x}}{1 + \frac{1}{x}}\right)^x \).
Then, taking the logarithm:
\[ \ln y = x \ln\left(\frac{1 - \frac{1}{x}}{1 + \frac{1}{x}}\right) \]

### Step 5: Simplify the logarithm
Now we need to simplify \( \ln\left(\frac{1 - \frac{1}{x}}{1 + \frac{1}{x}}\right) \):
\[ \ln\left(\frac{1 - \frac{1}{x}}{1 + \frac{1}{x}}\right) = \ln(1 - \frac{1}{x}) - \ln(1 + \frac{1}{x}) \]
Using the Taylor expansion for \( \ln(1 + u) \approx u \) when \( u \) is small, we have:
\[ \ln(1 - \frac{1}{x}) \approx -\frac{1}{x} \quad \text{and} \quad \ln(1 + \frac{1}{x}) \approx \frac{1}{x} \]
Thus,
\[ \ln\left(\frac{1 - \frac{1}{x}}{1 + \frac{1}{x}}\right) \approx -\frac{1}{x} - \frac{1}{x} = -\frac{2}{x} \]

### Step 6: Substitute back into the limit
Now substituting back:
\[ \ln y \approx x \left(-\frac{2}{x}\right) = -2 \]
Thus,
\[ y \approx e^{-2} \]

### Step 7: Conclude the limit
Therefore, we conclude that:
\[ \lim_{x \to \infty} \left(\frac{x-1}{x+1}\right)^x = e^{-2} \]

### Final Answer
The value of the limit is \( e^{-2} \).

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