009A Sample Midterm 1, Problem 2

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.

Foundations:
1. Left hand/right hand limits
2. Definition of limit in terms of right and left
3. Definition of continuous

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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