Limits of a Transfer Function as s approaches infinity

Question

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.

Answer 1

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 $0$.

Answer 2

write your term as $$\frac{s^4(3+\frac{108}{s^2})}{s^4(1+\frac{2.344}{s}+\frac{1.824}{s^2}+\frac{0.987}{s^4})}$$

Answer 3

Observe that, by dividing numerator and denominator by $s^4$, one gets $$ \frac{3s^2(s^2+36)}{s^4 + 2.344s^3 + 1.824s^2 + 0.987}=\frac{3+\frac{108}{s^2}}{1 + \frac{2.344}{s}+ \frac{1.824}{s^2} + \frac{0.987}{s^4}} $$ then let $s \to \infty$.