If you were to evaluate the limit as $s$ goes to infinity of the transfer function:
$$T(s) = {3s^2\left(s^2+36\right) \over s^4 + 2.344s^3 + 1.824s^2 + 0.987}$$
Why is the answer $3$? Is it due to the numerator having an $s$ term with a power of $4$ with $3$ as the coefficient, and the denominator has an $s$ term with power $4$ with a coefficient of $1$? Do you just ignore all the other terms?
EDIT: I am now trying to evaluate the limit as $s$ goes to zero.
It has been suggested to directly substitute $s = 0$ into the Transfer Function.
$$T(s) = {3(0)^2\left((0)^2+36\right) \over (0)^4 + 2.344(0)^3 + 1.824(0)^2 + 0.987}$$
However the answer is $$=3(s)^2 {36\over 0.987}$$
Why is there still an s term in the answer after the substitution.
EDIT 2:
Please refer to original question (1.9)(Ignore the second T(s) shown – the one with $s^5$, it is for another question) and solution provided. The solution provided may be wrong.
Best Answer
Since the rational function has a denominator and numerator of same degrees, the limit as $s \to \infty$ is the quotient of the numerator's leading coefficient and the denominator's leading coefficient. This is because as you increase towards $\infty$, terms other than leading one become negligible to the final result.
And thusly, if the degree of the numerator is larger, the limit approaches $\infty$. If the degree of the denominator is larger then the limit approaches $0$.