`lim_(x to 1) (x^(2) + x - 2)/(x^(2) - 1)` is equal to

Options

A. 0
B. `-1`
C. 2
D. `(3)/(2)`

Answer

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

### Step 1: Factor the numerator and denominator
First, we need to factor both the numerator \( x^2 + x - 2 \) and the denominator \( x^2 - 1 \).

- **Numerator**:
The expression \( x^2 + x - 2 \) can be factored by finding two numbers that multiply to \(-2\) (the constant term) and add to \(1\) (the coefficient of \(x\)). The numbers \(2\) and \(-1\) satisfy this condition. Thus, we can factor it as:
\[ x^2 + x - 2 = (x - 1)(x + 2) \]

- **Denominator**:
The expression \( x^2 - 1 \) is a difference of squares and can be factored as:
\[ x^2 - 1 = (x - 1)(x + 1) \]

### Step 2: Rewrite the limit with the factored forms
Now we can rewrite the limit using the factored forms:
\[ \lim_{x \to 1} \frac{(x - 1)(x + 2)}{(x - 1)(x + 1)} \]

### Step 3: Cancel the common factors
Since \( (x - 1) \) is a common factor in both the numerator and the denominator, we can cancel it out (as long as \( x \neq 1 \)):
\[ \lim_{x \to 1} \frac{x + 2}{x + 1} \]

### Step 4: Substitute \( x = 1 \)
Now we can directly substitute \( x = 1 \) into the simplified expression:
\[ \frac{1 + 2}{1 + 1} = \frac{3}{2} \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to 1} \frac{x^2 + x - 2}{x^2 - 1} = \frac{3}{2} \]

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