2. a population of birds (in millions) t years from now is estimated by the following function…

2. a population of birds (in millions) t years from now is estimated by the following function: $p(t)=\frac{30t^{2}+125t + 223}{4.5t^{2}+2t + 40}$. what will the population of the birds be in the long run?

2. a population of birds (in millions) t years from now is estimated by the following function: $p(t)=\frac{30t^{2}+125t + 223}{4.5t^{2}+2t + 40}$. what will the population of the birds be in the long run?

Answer

Explanation:

Step1: Find the limit as t approaches infinity

We need to find $\lim_{t\rightarrow\infty}P(t)=\lim_{t\rightarrow\infty}\frac{30t^{2}+125t + 223}{4.5t^{2}+2t + 40}$.

Step2: Divide numerator and denominator by $t^{2}$

$\lim_{t\rightarrow\infty}\frac{30t^{2}/t^{2}+125t/t^{2}+223/t^{2}}{4.5t^{2}/t^{2}+2t/t^{2}+40/t^{2}}=\lim_{t\rightarrow\infty}\frac{30+\frac{125}{t}+\frac{223}{t^{2}}}{4.5+\frac{2}{t}+\frac{40}{t^{2}}}$.

Step3: Evaluate the limit of each term

As $t\rightarrow\infty$, $\frac{125}{t}\rightarrow0$, $\frac{223}{t^{2}}\rightarrow0$, $\frac{2}{t}\rightarrow0$ and $\frac{40}{t^{2}}\rightarrow0$. So we have $\frac{30 + 0+0}{4.5+0 + 0}$.

Step4: Calculate the final result

$\frac{30}{4.5}=\frac{300}{45}=\frac{20}{3}\approx6.67$.

Answer:

$\frac{20}{3}$ (or approximately $6.67$ million)