# 007A Sample Midterm 1, Problem 2 Detailed Solution

Consider the following function  ${\displaystyle f:}$

${\displaystyle f(x)=\left\{{\begin{array}{lr}x^{2}&{\text{if }}x<1\\{\sqrt {x}}&{\text{if }}x\geq 1\end{array}}\right.}$

(a) Find  ${\displaystyle \lim _{x\rightarrow 1^{-}}f(x).}$

(b) Find  ${\displaystyle \lim _{x\rightarrow 1^{+}}f(x).}$

(c) Find  ${\displaystyle \lim _{x\rightarrow 1}f(x).}$

(d) Is  ${\displaystyle f}$  continuous at  ${\displaystyle x=1?}$  Briefly explain.

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

Solution:

(a)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 1^{-}}f(x)}&=&\displaystyle {\lim _{x\rightarrow 1^{-}}x^{2}}\\&&\\&=&\displaystyle {1^{2}}\\&&\\&=&\displaystyle {1.}\\\end{array}}}$

(b)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 1^{+}}f(x)}&=&\displaystyle {\lim _{x\rightarrow 1^{+}}{\sqrt {x}}}\\&&\\&=&\displaystyle {\sqrt {1}}\\&&\\&=&\displaystyle {1.}\\\end{array}}}$

(c)

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

(d)

Step 1:
From (c), we have
${\displaystyle \lim _{x\rightarrow 1}f(x)=1.}$
Also,
${\displaystyle f(1)={\sqrt {1}}=1.}$
Step 2:
Since
${\displaystyle \lim _{x\rightarrow 1}f(x)=f(1),}$
${\displaystyle f(x)}$  is continuous at  ${\displaystyle x=1.}$

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