Question

let (f(x)=\begin{cases}x^{2}+2&\text{if }x < 2\\(x - 3)^{2}&\text{if }xgeq2end{cases}) (a) find the following limits. (if an answer does not exist, enter dne.) (lim_{x
ightarrow2^{-}}f(x)=) (lim_{x
ightarrow2^{+}}f(x)=) (b) does (lim_{x
ightarrow2}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 $f(x)=x^{2}+2$. Substitute $x = 2$ into $x^{2}+2$:
$2^{2}+2=4 + 2=6$.

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 $(x - 3)^{2}$:
$(2 - 3)^{2}=(-1)^{2}=1$.

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 $6
eq1$, $\lim_{x
ightarrow2}f(x)$ does not exist.

Step4: Sketch the graph

The graph of $y = x^{2}+2$ for $x<2$ is a parabola opening upwards with vertex at $(0, 2)$. The graph of $y=(x - 3)^{2}$ for $x\geq2$ is a parabola opening upwards with vertex at $(3,0)$. At $x = 2$, the left - hand part of the graph approaches $y = 6$ and the right - hand part of the graph has the value $y = 1$.

Answer:

(a) $\lim_{x
ightarrow2^{-}}f(x)=6$
$\lim_{x
ightarrow2^{+}}f(x)=1$
(b) No
(c) The graph of $y=x^{2}+2$ for $x < 2$ and $y=(x - 3)^{2}$ for $x\geq2$ has a break at $x = 2$, with the left - hand side approaching $y = 6$ and the right - hand side having value $y = 1$.