# 007A Sample Midterm 3, Problem 2 Detailed Solution

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Suppose the size of a population at time  $t$ is given by

$N(t)={\frac {1000t}{5+t}},~t\geq 0.$ (a) Determine the size of the population as  $t\rightarrow \infty .$ We call this the limiting population size.

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

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

(a)

Step 1:
We have
$\lim _{t\rightarrow \infty }N(t)=\lim _{t\rightarrow \infty }{\frac {1000t}{5+t}}.$ Step 2:
Using the Background Information, we have
${\begin{array}{rcl}\displaystyle {\lim _{t\rightarrow \infty }N(t)}&=&\displaystyle {\frac {1000}{1}}\\&&\\&=&\displaystyle {1000.}\end{array}}$ (b)
We have
${\begin{array}{rcl}\displaystyle {N(5)}&=&\displaystyle {\frac {1000(5)}{5+5}}\\&&\\&=&\displaystyle {\frac {1000(5)}{10}}\\&&\\&=&\displaystyle {100(5)}\\&&\\&=&\displaystyle {500}\\&&\\&=&\displaystyle {{\frac {1000}{2}}.}\end{array}}$ (a)     $1000$ (b)     $N(5)=500$ 