009A Sample Midterm 1, Problem 2

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

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

Foundations:
1. If  $\lim _{x\rightarrow a^{-}}f(x)=\lim _{x\rightarrow a^{+}}f(x)=c,$
then  $\lim _{x\rightarrow a}f(x)=c.$
2.  $f(x)$  is continuous at  $x=a$  if
$\lim _{x\rightarrow a^{+}}f(x)=\lim _{x\rightarrow a^{-}}f(x)=f(a).$

Solution:

(a)

Step 1:
Notice that we are calculating a left hand limit.
Thus, we are looking at values of  $x$  that are smaller than  $1.$
Using the definition of  $f(x),$  we have
$\lim _{x\rightarrow 1^{-}}f(x)=\lim _{x\rightarrow 1^{-}}x^{2}.$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 1^{-}}f(x)}&=&\displaystyle {\lim _{x\rightarrow 1^{-}}x^{2}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 1}x^{2}}\\&&\\&=&\displaystyle {1^{2}}\\&&\\&=&\displaystyle {1.}\\\end{array}}$

(b)

Step 1:
Notice that we are calculating a right hand limit.
Thus, we are looking at values of  $x$  that are bigger than  $1.$
Using the definition of  $f(x),$  we have
$\lim _{x\rightarrow 1^{+}}f(x)=\lim _{x\rightarrow 1^{+}}{\sqrt {x}}.$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 1^{+}}f(x)}&=&\displaystyle {\lim _{x\rightarrow 1^{+}}{\sqrt {x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 1}{\sqrt {x}}}\\&&\\&=&\displaystyle {\sqrt {1}}\\&&\\&=&\displaystyle {1.}\\\end{array}}$

(c)

Step 1:
From (a) and (b), we have
$\lim _{x\rightarrow 1^{-}}f(x)=1$
and
$\lim _{x\rightarrow 1^{+}}f(x)=1.$

Step 2:
Since
$\lim _{x\rightarrow 1^{-}}f(x)=\lim _{x\rightarrow 1^{+}}f(x)=1,$
we have
$\lim _{x\rightarrow 1}f(x)=1.$

(d)

Step 1:
From (c), we have
$\lim _{x\rightarrow 1}f(x)=1.$
Also,
$f(1)={\sqrt {1}}=1.$

Step 2:
Since
$\lim _{x\rightarrow 1}f(x)=f(1),$
$f(x)$  is continuous at  $x=1.$

Final Answer:
(a)     $1$
(b)     $1$
(c)     $1$
(d)     $f(x)$  is continuous at  $x=1$  since  $\lim _{x\rightarrow 1}f(x)=f(1).$

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