Answer:
1600. This is a maximum limit of fish the pond can host.
Step-by-step explanation:
For [tex]\lim_{t \to \infty} N(t)[/tex], we must take the part of this function valid for high values, i.e., the second part, where [tex]t \geq 8\\[/tex].
[tex]\lim_{t \to \infty} N(t)=\lim_{t \to \infty} \frac{200+80t}{2+0.05t}[/tex]
Since we have two polynoms both in numerator and denominator, and both of them are of degree 1 (both linear), for high values of [tex]t[/tex], the main part of each polynom shall be the linear part, neglecting lower degree parts (in this case, constant terms):
[tex]\lim_{t \to \infty} \frac{200+80t}{2+0.05t}=\lim_{t \to \infty} \frac{80t}{0.05t}=\frac{80}{0.05} = 1600[/tex]
This means that the number of fish in a pond has an horizontal asymthote. In other words, there seems to be a natural limit for the number of fish that there will be in the pond as years pass. The maximum number of fish is actually 1600. With this function, no higher than this figure can be reached. This might imply limits in productivity.
The limit of the function is given by:
[tex]\lim_{t \rightarrow \infty} N(t) = 1600[/tex]
In the context of this problem, it means that as time goes to infinity, the population will tend to 1600, which is the carrying capacity of the pond.
The population after t years is modeled by:
[tex]N(t) = 25t + 150, 0 \leq t < 8[/tex]
[tex]N(t) = \frac{200 + 80t}{2 + 0.05t}, t \geq 8[/tex]
[tex]t \rightarrow \infty \geq 8[/tex], thus:
[tex]\lim_{t \rightarrow \infty} N(t) = \lim_{t \rightarrow \infty} \frac{200 + 80t}{2 + 0.05t} = \lim_{t \rightarrow \infty} \frac{80t}{0.05t} = \lim_{t \rightarrow \infty} 1600 = 1600[/tex]
In the context of this problem, it means that as time goes to infinity, the population will tend to 1600, which is the carrying capacity of the pond.
A similar problem is given at https://brainly.com/question/16553497
