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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 1^-} f(x)=\lim_{x\rightarrow 1^-} x^2.}

Step 2:
Now, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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:

Step 2:

(c)

Step 1:

Step 2:

(d)

Step 1:

Step 2:

Final Answer:
(a)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1}
(b)
(c)
(d)

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