Evaluate the following limit:
`lim_(x->1)(1/(x^2+x-2)-x/(x^3-1))`

Options

A. `1/3`
B. `-1/3`
C. `-1/9`
D. `1/9`

Answer

To evaluate the limit

\[ \lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right), \]

we will follow these steps:

### Step 1: Substitute \( x = 1 \)

First, we will substitute \( x = 1 \) into the expression to check if we can directly evaluate the limit.

\[ \frac{1}{1^2 + 1 - 2} - \frac{1}{1^3 - 1} = \frac{1}{1 + 1 - 2} - \frac{1}{1 - 1} = \frac{1}{0} - \frac{1}{0}. \]

This gives us the form \(\frac{1}{0} - \frac{1}{0}\), which is an indeterminate form (infinity minus infinity).

**Hint:** Check if substituting directly gives an indeterminate form, which indicates further simplification is needed.

### Step 2: Factor the denominators

Next, we will factor the denominators to simplify the expression.

The first denominator can be factored as follows:

\[ x^2 + x - 2 = (x - 1)(x + 2). \]

The second denominator can be factored using the difference of cubes:

\[ x^3 - 1 = (x - 1)(x^2 + x + 1). \]

### Step 3: Rewrite the limit

Now we can rewrite the limit using the factored forms:

\[ \lim_{x \to 1} \left( \frac{1}{(x - 1)(x + 2)} - \frac{x}{(x - 1)(x^2 + x + 1)} \right). \]

### Step 4: Combine the fractions

To combine the fractions, we will find a common denominator, which is \((x - 1)(x + 2)(x^2 + x + 1)\):

\[ \lim_{x \to 1} \left( \frac{(x^2 + x + 1) - x(x + 2)}{(x - 1)(x + 2)(x^2 + x + 1)} \right). \]

### Step 5: Simplify the numerator

Now, simplify the numerator:

\[ (x^2 + x + 1) - x(x + 2) = x^2 + x + 1 - (x^2 + 2x) = x^2 + x + 1 - x^2 - 2x = -x + 1. \]

Thus, we can rewrite the limit as:

\[ \lim_{x \to 1} \frac{-x + 1}{(x - 1)(x + 2)(x^2 + x + 1)}. \]

### Step 6: Factor out \(-1\)

Notice that \(-x + 1 = -(x - 1)\). So we can factor that out:

\[ \lim_{x \to 1} \frac{-(x - 1)}{(x - 1)(x + 2)(x^2 + x + 1)}. \]

### Step 7: Cancel out the common terms

Now we can cancel \((x - 1)\) from the numerator and denominator:

\[ \lim_{x \to 1} \frac{-1}{(x + 2)(x^2 + x + 1)}. \]

### Step 8: Substitute \( x = 1 \) again

Now we can substitute \( x = 1 \):

\[ \frac{-1}{(1 + 2)(1^2 + 1 + 1)} = \frac{-1}{3 \cdot 3} = \frac{-1}{9}. \]

### Final Answer

Thus, the limit evaluates to:

\[ \lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right) = -\frac{1}{9}. \]

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