Evaluate the following limits :
`lim_(x to 1)[(x-2)/(x^(2)-x)-1/(x^(3)-3x^(2)+2x)]`

Answer

To evaluate the limit

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

we will follow these steps:

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

First, we substitute \(x = 1\) into the expression to check the form of the limit.

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

Both terms approach \(-\infty\), indicating an indeterminate form.

**Hint:** Always check the form of the limit by substituting the value into the expression.

### Step 2: Simplify the expression

Next, we simplify the expression. The first term can be rewritten as:

\[ \frac{x-2}{x^2 - x} = \frac{x-2}{x(x-1)}. \]

The second term can be factored:

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

Thus, the expression becomes:

\[ \frac{x-2}{x(x-1)} - \frac{1}{x(x-1)(x-2)}. \]

**Hint:** Factor polynomials to simplify expressions before taking limits.

### Step 3: Combine the fractions

Now we combine the two fractions:

\[ \frac{(x-2)(x-2) - 1}{x(x-1)(x-2)} = \frac{(x-2)^2 - 1}{x(x-1)(x-2)}. \]

### Step 4: Expand the numerator

Next, we expand the numerator:

\[ (x-2)^2 - 1 = x^2 - 4x + 4 - 1 = x^2 - 4x + 3. \]

So now we have:

\[ \frac{x^2 - 4x + 3}{x(x-1)(x-2)}. \]

### Step 5: Factor the numerator

The numerator can be factored as:

\[ x^2 - 4x + 3 = (x-3)(x-1). \]

Thus, the limit expression simplifies to:

\[ \frac{(x-3)(x-1)}{x(x-1)(x-2)}. \]

### Step 6: Cancel common factors

We can cancel the \((x-1)\) terms:

\[ \frac{x-3}{x(x-2)}. \]

### Step 7: Take the limit as \(x \to 1\)

Now we substitute \(x = 1\):

\[ \lim_{x \to 1} \frac{x-3}{x(x-2)} = \frac{1-3}{1(1-2)} = \frac{-2}{1 \cdot -1} = 2. \]

### Final Answer

Thus, the limit evaluates to:

\[ \boxed{2}. \]

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