Let \( f(x) = \frac{x^2 - 1}{|x| - 1} \). Then the value of \( \lim_{x \to -1} f(x) \) is?
Step 1: Understand the Function
First, I need to understand the function \( f(x) = \frac{x^2 - 1}{|x| - 1} \).
The numerator is \( x^2 - 1 \).
The denominator is \( |x| - 1 \).
I know that \( |x| \) is the absolute value of \( x \), so \( |x| = x \) if \( x \geq 0 \) and \( |x| = -x \) if \( x < 0 \).
Since we're taking the limit as \( x \) approaches -1, which is less than 0, \( |x| = -x \) in this case.
Step 2: Simplify the Function
Given that \( x \) is approaching -1, \( |x| = -x \). So, the denominator becomes: \[ |x| - 1 = -x - 1 \] Now, the function \( f(x) \) can be rewritten as: \[ f(x) = \frac{x^2 - 1}{-x - 1} \] I can factor the numerator: \[ x^2 - 1 = (x - 1)(x + 1) \] So, the function becomes: \[ f(x) = \frac{(x - 1)(x + 1)}{-x - 1} \] I notice that \( -x - 1 = -(x + 1) \), so: \[ f(x) = \frac{(x - 1)(x + 1)}{-(x + 1)} = \frac{(x - 1)}{-1} = -(x - 1) = -x + 1 \] Now, I need to find: \[ \lim_{x \to -1} f(x) = \lim_{x \to -1} (-x + 1) \] This is a simple linear function, so I can directly substitute \( x = -1 \): \[ \lim_{x \to -1} (-x + 1) = -(-1) + 1 = 1 + 1 = 2 \]