Question

define f(2) in a way that extends f(s) = \frac{s^{3}-8}{s^{2}-4} to be continuous at s = 2. f(2)= (type an integer or a simplified fraction.)

Explanation:

Step1: Factor the numerator and denominator

The numerator $s^{3}-8=(s - 2)(s^{2}+2s + 4)$ using the formula $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$ with $a=s$ and $b = 2$. The denominator $s^{2}-4=(s - 2)(s + 2)$ using the difference - of - squares formula $a^{2}-b^{2}=(a - b)(a + b)$ with $a=s$ and $b = 2$. So, $f(s)=\frac{(s - 2)(s^{2}+2s + 4)}{(s - 2)(s + 2)}$.

Step2: Simplify the function

Cancel out the common factor $(s - 2)$ (for $s
eq2$), we get $f(s)=\frac{s^{2}+2s + 4}{s + 2}$ for $s
eq2$.

Step3: Find the limit as $s\to2$

To make the function continuous at $s = 2$, we find $\lim_{s\to2}f(s)$. Substitute $s = 2$ into $\frac{s^{2}+2s + 4}{s + 2}$. We have $\frac{2^{2}+2\times2 + 4}{2+2}=\frac{4 + 4+4}{4}=\frac{12}{4}=3$.

Answer:

$3$