Question

Define f(10) in a way that extends f(s)=(s^3-1000)/(s^2-100) to be continuous at s=10....help, I'm stuck!

Answer 1

We want $f$ to be continuous at $s=10,$ so the following limit must exist:

$L=\underset{s\to 10}{\mathrm{\ell im}}~f(s)\\\\\\ L=\underset{s\to 10}{\mathrm{\ell im}}~\dfrac{s^3-1000}{s^2-100}$


Now there's an indeterminated form "0/0". So we can factor out (s-10) in the fraction terms (through polynomial division):

$L=\underset{s\to 10}{\mathrm{\ell im}}~\dfrac{(s-10)(s^2+10s+100)}{(s-10)(s+10)}\\\\\\ L=\underset{s\to 10}{\mathrm{\ell im}}~\dfrac{s^2+10s+100}{s+10}\\\\\\ L=\dfrac{10^2+10\cdot 10+100}{10+10}\\\\\\ L=\dfrac{100+100+100}{10+10}\\\\\\ L=\dfrac{300}{20}\\\\\\ L=15$

____________

So if we define

$f(10)=15,$

then the limit of $f$ as $s$ approaches 10 is equal to f(10), which means that $f$ is continuous at $s=10.$