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

Answer

To evaluate the limit

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

we will follow these steps:

### Step 1: Analyze the limit

As \( x \to 1 \), both terms in the limit approach infinity. Specifically, \( \frac{1}{x-1} \) approaches infinity and the denominator \( x^3 - 3x^2 + 2 \) also approaches zero. Therefore, we can expect that we may need to simplify the expression.

### Step 2: Factor the denominator

First, let's factor the polynomial in the denominator:

\[ x^3 - 3x^2 + 2. \]

To factor this polynomial, we can try to find its roots. We can use the Rational Root Theorem or synthetic division. Testing \( x = 1 \):

\[ 1^3 - 3(1^2) + 2 = 1 - 3 + 2 = 0. \]

So, \( x - 1 \) is a factor. Now we can perform polynomial long division or synthetic division to factor \( x^3 - 3x^2 + 2 \):

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

### Step 3: Rewrite the limit

Now we can rewrite the limit:

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

### Step 4: Combine the fractions

We can combine the two fractions over a common denominator:

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

### Step 5: Simplify the numerator

Now simplify the numerator:

\[ x^2 - 2x - 2 - 3(x - 2) = x^2 - 2x - 2 - 3x + 6 = x^2 - 5x + 4. \]

### Step 6: Factor the numerator

Next, we can factor \( x^2 - 5x + 4 \):

\[ x^2 - 5x + 4 = (x - 1)(x - 4). \]

### Step 7: Substitute back into the limit

Substituting back, we have:

\[ \lim_{x \to 1} \frac{(x - 1)(x - 4)}{(x - 1)(x^2 - 2x - 2)}. \]

### Step 8: Cancel common factors

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

\[ \lim_{x \to 1} \frac{x - 4}{x^2 - 2x - 2}. \]

### Step 9: Evaluate the limit

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

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

Thus, the limit evaluates to:

\[ \boxed{1}. \]