As x gets closer to 2, `(x^3-8)/(x^2-4)` approaches

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the limit problem \(\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\), we can follow these steps:

### Step 1: Identify the limit
We start with the expression:
\[ \lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4} \]

### Step 2: Substitute \(x = 2\)
If we directly substitute \(x = 2\):
\[ \frac{2^3 - 8}{2^2 - 4} = \frac{8 - 8}{4 - 4} = \frac{0}{0} \]
This gives us an indeterminate form \(0/0\), so we need to simplify the expression.

### Step 3: Factor the numerator and denominator
We can factor both the numerator and the denominator.

- The numerator \(x^3 - 8\) can be factored using the difference of cubes:
\[ x^3 - 8 = (x - 2)(x^2 + 2x + 4) \]
- The denominator \(x^2 - 4\) can be factored using the difference of squares:
\[ x^2 - 4 = (x - 2)(x + 2) \]

### Step 4: Rewrite the limit
Now we can rewrite the limit with the factored forms:
\[ \lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)} \]

### Step 5: Cancel common factors
Since \(x - 2\) is a common factor in both the numerator and the denominator, we can cancel it (as long as \(x \neq 2\)):
\[ \lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 2} \]

### Step 6: Substitute \(x = 2\) again
Now we can substitute \(x = 2\) into the simplified expression:
\[ \frac{2^2 + 2(2) + 4}{2 + 2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3 \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4} = 3 \]