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 limit value
First, we substitute \( x = 2 \) into the numerator and denominator to check if we get an indeterminate form.

**Numerator:**
\[ 2^3 - 3(2^2) + 4 = 8 - 12 + 4 = 0. \]

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

Since both the numerator and denominator evaluate to 0, we have the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator separately.

**Differentiate the numerator:**
\[ \frac{d}{dx}(x^3 - 3x^2 + 4) = 3x^2 - 6x. \]

**Differentiate the denominator:**
\[ \frac{d}{dx}(x^4 - 8x^2 + 16) = 4x^3 - 16x. \]

### Step 3: Rewrite the limit
Now we rewrite the limit using the derivatives we found:

\[ \lim_{x \to 2} \frac{3x^2 - 6x}{4x^3 - 16x}. \]

### Step 4: Substitute the limit value again
Now we substitute \( x = 2 \) into the new numerator and denominator.

**New Numerator:**
\[ 3(2^2) - 6(2) = 3(4) - 12 = 12 - 12 = 0. \]

**New Denominator:**
\[ 4(2^3) - 16(2) = 4(8) - 32 = 32 - 32 = 0. \]

We still have the indeterminate form \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 5: Differentiate again
**Differentiate the numerator again:**
\[ \frac{d}{dx}(3x^2 - 6x) = 6x - 6. \]

**Differentiate the denominator again:**
\[ \frac{d}{dx}(4x^3 - 16x) = 12x^2 - 16. \]

### Step 6: Rewrite the limit again
Now we rewrite the limit using the new derivatives:

\[ \lim_{x \to 2} \frac{6x - 6}{12x^2 - 16}. \]

### Step 7: Substitute the limit value one more time
Now we substitute \( x = 2 \):

**New Numerator:**
\[ 6(2) - 6 = 12 - 6 = 6. \]

**New Denominator:**
\[ 12(2^2) - 16 = 12(4) - 16 = 48 - 16 = 32. \]

### Step 8: Calculate the limit
Now we can calculate the limit:

\[ \lim_{x \to 2} \frac{6}{32} = \frac{3}{16}. \]

Thus, the final answer is:

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