Evaluate following limits :
`lim_( x to -1)(x^(2)-1)/(x+1)`

Answer

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

### Step 1: Identify the form of the limit
First, we substitute \( x = -1 \) into the expression:
\[ \frac{(-1)^2 - 1}{-1 + 1} = \frac{1 - 1}{0} = \frac{0}{0} \]
This is an indeterminate form, which means we need to simplify the expression.

### Step 2: Factor the numerator
We recognize that \( x^2 - 1 \) can be factored as a difference of squares:
\[ x^2 - 1 = (x - 1)(x + 1) \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to -1} \frac{(x - 1)(x + 1)}{x + 1} \]

### Step 3: Cancel common factors
Since \( x + 1 \) is in both the numerator and the denominator, we can cancel it (as long as \( x \neq -1 \)):
\[ \lim_{x \to -1} (x - 1) \]

### Step 4: Substitute the limit value
Now we can directly substitute \( x = -1 \) into the simplified expression:
\[ -1 - 1 = -2 \]

### Conclusion
Thus, the limit evaluates to:
\[ \lim_{x \to -1} \frac{x^2 - 1}{x + 1} = -2 \]