Question

$$lim_{x ightarrow2}\frac{x^{3}-8}{x - 2}$$

Explanation:

Step1: Factor the numerator

We know that $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$. Here $a=x$ and $b = 2$, so $x^{3}-8=(x - 2)(x^{2}+2x + 4)$. Then the limit becomes $\lim_{x
ightarrow2}\frac{(x - 2)(x^{2}+2x + 4)}{x - 2}$.

Step2: Simplify the expression

Cancel out the common factor $(x - 2)$ in the numerator and denominator. We get $\lim_{x
ightarrow2}(x^{2}+2x + 4)$.

Step3: Substitute $x = 2$

Substitute $x=2$ into $x^{2}+2x + 4$. So $2^{2}+2\times2+4=4 + 4+4=12$.

Answer:

$12$