Evaluate the following limits :
`Lim_(x to oo) ((x-1)/(x+1))^(2)`

Answer

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

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

### Step 2: Simplify the fraction
Next, we 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 3: Substitute the limit
Now we substitute this back into our limit:
\[ L = \lim_{x \to \infty} \left( \frac{1 - \frac{1}{x}}{1 + \frac{1}{x}} \right)^2 \]

### Step 4: Evaluate the limit
As \( x \) approaches infinity, \( \frac{1}{x} \) approaches 0. Therefore:
\[ L = \left( \frac{1 - 0}{1 + 0} \right)^2 = \left( \frac{1}{1} \right)^2 = 1^2 = 1 \]

### Final Answer
Thus, the limit is:
\[ \boxed{1} \]