Question

let
$f(x)=\

$$\begin{cases}x^{2}+1&\\text{if }x < 2\\\\(x - 3)^{2}&\\text{if }x\\geq2\\end{cases}$$

$
(a) find the following limits. (if an answer does not exist, enter dne.)
$\lim_{x\to2^{-}}f(x)=$
$\lim_{x\to2^{+}}f(x)=$
(b) does $\lim_{x\to2}f(x)$ exist?
yes
no
(c) sketch the graph of $f$.

Explanation:

Step1: Find left - hand limit

For $\lim_{x
ightarrow2^{-}}f(x)$, since $x
ightarrow2^{-}$ means $x < 2$, we use $f(x)=x^{2}+1$. Substitute $x = 2$ into $x^{2}+1$. So, $\lim_{x
ightarrow2^{-}}f(x)=2^{2}+1=5$.

Step2: Find right - hand limit

For $\lim_{x
ightarrow2^{+}}f(x)$, since $x
ightarrow2^{+}$ means $x\geq2$, we use $f(x)=(x - 3)^{2}$. Substitute $x = 2$ into $(x - 3)^{2}$. So, $\lim_{x
ightarrow2^{+}}f(x)=(2 - 3)^{2}=1$.

Step3: Determine if the limit exists

The limit $\lim_{x
ightarrow2}f(x)$ exists if and only if $\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{+}}f(x)$. Since $5
eq1$, $\lim_{x
ightarrow2}f(x)$ does not exist.

Answer:

(a) $\lim_{x
ightarrow2^{-}}f(x)=5$
$\lim_{x
ightarrow2^{+}}f(x)=1$
(b) No
(c) To sketch the graph:

• For $y=x^{2}+1$ when $x < 2$, it is a parabola opening upwards with vertex at $(0,1)$. The point $(2,5)$ is a hole (since $x<2$).
• For $y=(x - 3)^{2}$ when $x\geq2$, it is a parabola opening upwards with vertex at $(3,0)$. The point $(2,1)$ is a solid - dot.