We are asked to find the limit: $lim_{x \to 2} \frac{x^3 - 8}{\sqrt{x+2} - 2}$.

1. Problem Description

We are asked to find the limit:

$lim_{x \to 2} \frac{x^3 - 8}{\sqrt{x+2} - 2}$.

2. Solution Steps

First, we note that if we directly substitute $x=2$ into the expression, we get $\frac{2^3 - 8}{\sqrt{2+2} - 2} = \frac{8-8}{\sqrt{4} - 2} = \frac{0}{0}$, which is an indeterminate form.

We can factor the numerator using the difference of cubes formula:

$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.

So, $x^3 - 8 = x^3 - 2^3 = (x-2)(x^2 + 2x + 4)$.

Next, we can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{x+2} + 2$:

$\frac{x^3 - 8}{\sqrt{x+2} - 2} = \frac{(x-2)(x^2 + 2x + 4)}{\sqrt{x+2} - 2} \cdot \frac{\sqrt{x+2} + 2}{\sqrt{x+2} + 2} = \frac{(x-2)(x^2 + 2x + 4)(\sqrt{x+2} + 2)}{(\sqrt{x+2})^2 - 2^2} = \frac{(x-2)(x^2 + 2x + 4)(\sqrt{x+2} + 2)}{x+2 - 4} = \frac{(x-2)(x^2 + 2x + 4)(\sqrt{x+2} + 2)}{x-2}$.

Now, we can cancel the $(x-2)$ term in the numerator and denominator, since we are taking the limit as $x$ approaches 2, not when $x$ is equal to

2.

$\frac{(x-2)(x^2 + 2x + 4)(\sqrt{x+2} + 2)}{x-2} = (x^2 + 2x + 4)(\sqrt{x+2} + 2)$ for $x \neq 2$.

Now, we can take the limit as $x \to 2$:

$lim_{x \to 2} (x^2 + 2x + 4)(\sqrt{x+2} + 2) = (2^2 + 2(2) + 4)(\sqrt{2+2} + 2) = (4 + 4 + 4)(\sqrt{4} + 2) = (12)(2+2) = (12)(4) = 48$.

3. Final Answer

48