007A Sample Midterm 1, Problem 2 Detailed Solution

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Consider the following function  $f:$ $f(x)=\left\{{\begin{array}{lr}x^{2}&{\text{if }}x<1\\{\sqrt {x}}&{\text{if }}x\geq 1\end{array}}\right.$ (a) Find  $\lim _{x\rightarrow 1^{-}}f(x).$ (b) Find  $\lim _{x\rightarrow 1^{+}}f(x).$ (c) Find  $\lim _{x\rightarrow 1}f(x).$ (d) Is  $f$ continuous at  $x=1?$ Briefly explain.

Background Information:
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

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

${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 1^{+}}f(x)}&=&\displaystyle {\lim _{x\rightarrow 1^{+}}{\sqrt {x}}}\\&&\\&=&\displaystyle {\sqrt {1}}\\&&\\&=&\displaystyle {1.}\\\end{array}}$ (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  $x=1.$ (a)     $1$ (b)     $1$ (c)     $1$ (d)     $f(x)$ is continuous at  $x=1$ since  $\lim _{x\rightarrow 1}f(x)=f(1).$ 