# 009A Sample Midterm 1, Problem 2 Detailed Solution

Suppose the size of a population at time  ${\displaystyle t}$  is given by

${\displaystyle N(t)={\frac {1000t}{5+t}},~t\geq 0.}$

(a) Determine the size of the population as  ${\displaystyle t\rightarrow \infty .}$  We call this the limiting population size.

(b) Show that at time  ${\displaystyle t=5,}$  the size of the population is half its limiting size.

Background Information:
Recall that
${\displaystyle \lim _{x\rightarrow \infty }{\frac {a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{0}}{b_{n}x^{n}+b_{n-1}x^{n-1}+\cdots +b_{0}}}={\frac {a_{n}}{b_{n}}}}$
provided  ${\displaystyle a_{n}\neq 0}$  and  ${\displaystyle b_{n}\neq 0.}$

Solution:

(a)

Step 1:
We have
${\displaystyle \lim _{t\rightarrow \infty }N(t)=\lim _{t\rightarrow \infty }{\frac {1000t}{5+t}}.}$
Step 2:
Using the Background Information, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{t\rightarrow \infty }N(t)}&=&\displaystyle {\frac {1000}{1}}\\&&\\&=&\displaystyle {1000.}\end{array}}}$

(b)
We have
${\displaystyle {\begin{array}{rcl}\displaystyle {N(5)}&=&\displaystyle {\frac {1000(5)}{5+5}}\\&&\\&=&\displaystyle {\frac {1000(5)}{10}}\\&&\\&=&\displaystyle {100(5)}\\&&\\&=&\displaystyle {500}\\&&\\&=&\displaystyle {{\frac {1000}{2}}.}\end{array}}}$

(a)     ${\displaystyle 1000}$
(b)     ${\displaystyle N(5)=500}$