Verified Answer :

lukyo

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 \(s=10.\)

Factoring the expression, it is found that the definition of the function at s = 10 should be:

\(f(10) = 15\)

To make the function continuous at s = 10, we have to factor both the numerator and the denominator with relation to \(s + 10\), thus:

At the denominator:

\(s^2 - 100 = (s - 10)(s + 10)\)

At the numerator:

\((as^2 + bs + c)(s - 10) = s^3 - 1000\)

\(as^3 + (b - 10a)s^2 + (c - 10b)s - 10c = s^3 - 1000\)

Equaling both sides:

\(a = 1\)

\(b - 10a = 0 \rightarrow b = 10\)

\(-10c = -1000 \rightarrow = c = 100\)

Thus, the simplified function is:

\(f(s) = \frac{(s^2 + 10s + 100)(s - 10)}{(s + 10)(s - 10)} = \frac{s^2 + 10s + 100}{s + 10}\)

Considering s = 10:

\(f(10) = \frac{(10)^2 + 10(10) + 100}{10 + 10} = 15\)

Thus, the definition is:

\(f(10) = 15\)

A similar problem is given at https://brainly.com/question/12701640

Related Questions