Evaluate `lim_(x to 2) (x^(3) - 3x^(2) + 4)/(x^(4) - 8x^(2) + 16)`

Explanation

To evaluate the limit

\[ \lim_{x \to 2} \frac{x^3 - 3x^2 + 4}{x^4 - 8x^2 + 16}, \]

we will follow these steps:

### Step 1: Substitute the value of \(x\)
First, we substitute \(x = 2\) into the expression to check if we get a determinate form.

\[ \text{Numerator: } 2^3 - 3(2^2) + 4 = 8 - 12 + 4 = 0, \]

\[ \text{Denominator: } 2^4 - 8(2^2) + 16 = 16 - 32 + 16 = 0. \]

Both the numerator and the denominator evaluate to 0, which gives us the indeterminate form \(\frac{0}{0}\).

### Step 2: Factor the numerator and denominator
Since we have an indeterminate form, we need to factor both the numerator and the denominator.

**Numerator:**
The numerator \(x^3 - 3x^2 + 4\) can be factored or simplified. We can use polynomial long division or synthetic division to find the roots. Let's check if \(x = 2\) is a root:

Using synthetic division with root \(2\):

\[ \begin{array}{r|rrrr} 2 & 1 & -3 & 0 & 4 \\ & & 2 & -2 & -4 \\ \hline & 1 & -1 & -2 & 0 \\ \end{array} \]

This gives us:

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

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

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

Thus, the numerator becomes:

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

**Denominator:**
Now, we factor the denominator \(x^4 - 8x^2 + 16\). We can rewrite it as:

\[ x^4 - 8x^2 + 16 = (x^2 - 4)^2 = (x - 2)^2 (x + 2)^2. \]

### Step 3: Rewrite the limit
Now, we can rewrite the limit:

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

### Step 4: Cancel out the common factors
We can cancel \((x - 2)^2\) from the numerator and denominator:

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

### Step 5: Substitute \(x = 2\) again
Now we substitute \(x = 2\):

\[ \frac{2 + 1}{(2 + 2)^2} = \frac{3}{4^2} = \frac{3}{16}. \]

### Conclusion
Thus, the limit is:

\[ \lim_{x \to 2} \frac{x^3 - 3x^2 + 4}{x^4 - 8x^2 + 16} = \frac{3}{16}. \]