Question

let
f(x)=\begin{cases}x^{2}+3&\text{if }x < 1\\(x - 2)^{2}&\text{if }xgeq1end{cases}
(a) find the following limits. (if an answer does not exist, enter dne.)
(lim_{x
ightarrow1^{-}}f(x)=)
(lim_{x
ightarrow1^{+}}f(x)=)
(b) does (lim_{x
ightarrow1}f(x)) exist?
yes
no
(c) sketch the graph of (f).

Explanation:

Step1: Find left - hand limit

For $\lim_{x
ightarrow1^{-}}f(x)$, since $x
ightarrow1^{-}$ means $x < 1$, we use $f(x)=x^{2}+3$. Substitute $x = 1$ into $x^{2}+3$: $1^{2}+3=4$.

Step2: Find right - hand limit

For $\lim_{x
ightarrow1^{+}}f(x)$, since $x
ightarrow1^{+}$ means $x\geq1$, we use $f(x)=(x - 2)^{2}$. Substitute $x = 1$ into $(x - 2)^{2}$: $(1 - 2)^{2}=1$.

Step3: Determine if the limit exists

The limit $\lim_{x
ightarrow1}f(x)$ exists if and only if $\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x)$. Since $4
eq1$, $\lim_{x
ightarrow1}f(x)$ does not exist.

Answer:

(a) $\lim_{x
ightarrow1^{-}}f(x)=4$
$\lim_{x
ightarrow1^{+}}f(x)=1$
(b) No
(c) For $y = x^{2}+3,x<1$, it is a part of a parabola opening upwards with vertex at $(0,3)$ and an open - circle at $(1,4)$. For $y=(x - 2)^{2},x\geq1$, it is a part of a parabola opening upwards with vertex at $(2,0)$ and a closed - circle at $(1,1)$.