Evauate `lim_(xto1) (x^(4)-3x^(4)+2)/(x^(3)-5x^(2)+3x+1).`

Explanation

To evaluate the limit

\[ \lim_{x \to 1} \frac{x^4 - 3x^2 + 2}{x^3 - 5x^2 + 3x + 1}, \]

we will follow these steps:

### Step 1: Factor the Numerator

The numerator is \(x^4 - 3x^2 + 2\). We can substitute \(t = x^2\), so the expression becomes:

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

Next, we factor this quadratic:

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

Substituting back \(t = x^2\), we have:

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

Now, we can factor \(x^2 - 1\) further:

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

Thus, the numerator can be expressed as:

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

### Step 2: Factor the Denominator

Now, we need to factor the denominator \(x^3 - 5x^2 + 3x + 1\). We can use the Rational Root Theorem to find a root. Testing \(x = 1\):

\[ 1^3 - 5(1^2) + 3(1) + 1 = 1 - 5 + 3 + 1 = 0. \]

Thus, \(x = 1\) is a root. We can perform synthetic division of \(x^3 - 5x^2 + 3x + 1\) by \(x - 1\):

1 | 1 -5 3 1
| 1 -4 -1
-------------------
| 1 -4 -1 0

The quotient is \(x^2 - 4x - 1\). Therefore, we can write the denominator as:

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

### Step 3: Cancel Common Factors

Now we can rewrite the limit:

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

Since both the numerator and denominator have a common factor of \(x - 1\), we can cancel it out (as long as \(x \neq 1\)):

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

### Step 4: Substitute \(x = 1\)

Now we substitute \(x = 1\):

Numerator:

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

Denominator:

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

Thus, the limit becomes:

\[ \frac{-2}{-4} = \frac{1}{2}. \]

### Final Answer

The limit is:

\[ \lim_{x \to 1} \frac{x^4 - 3x^2 + 2}{x^3 - 5x^2 + 3x + 1} = \frac{1}{2}. \]