Question

use the function and graph to answer the questions about existence, limits, and continuity.
f(x) = \

$$\begin{cases}-x - 3, & - 5\\leq x < - 4 \\\\x + 5, & - 4 < x < - 2 \\\\-\frac{3}{2}x, & - 2 < x < 0 \\\\0, & 0\\leq x\\leq1 \\\\2x - 2, & 1 < x < 3 \\\\2, & x = 3 \\\\-2x + 10, & 3 < x < 5 \\\\1, & x = 5\\end{cases}$$

does f(-2) exist?
yes
no
does \lim_{x\to - 2}f(x) exist?
no
yes
does \lim_{x\to - 2}f(x) equal f(-2)?
no
yes

Explanation:

Step1: Check if f(-2) exists

The function is defined as \(f(x)=x + 5\) for \(-4

Step2: Check if \(\lim_{x

ightarrow - 2}f(x)\) exists
We find the left - hand limit \(\lim_{x
ightarrow - 2^{-}}f(x)\) and right - hand limit \(\lim_{x
ightarrow - 2^{+}}f(x)\).
For the left - hand limit, as \(x
ightarrow - 2^{-}\), \(f(x)=x + 5\), so \(\lim_{x
ightarrow - 2^{-}}f(x)=-2 + 5=3\).
For the right - hand limit, as \(x
ightarrow - 2^{+}\), \(f(x)=-\frac{3}{2}x\), so \(\lim_{x
ightarrow - 2^{+}}f(x)=-\frac{3}{2}\times(-2)=3\). Since \(\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{+}}f(x)=3\), \(\lim_{x
ightarrow - 2}f(x)\) exists.

Step3: Check if \(\lim_{x

ightarrow - 2}f(x)=f(-2)\)
Since \(f(-2)\) does not exist and \(\lim_{x
ightarrow - 2}f(x)=3\), \(\lim_{x
ightarrow - 2}f(x)
eq f(-2)\) (because \(f(-2)\) is undefined).

Answer:

Does \(f(-2)\) exist? No
Does \(\lim_{x
ightarrow - 2}f(x)\) exist? Yes
Does \(\lim_{x
ightarrow - 2}f(x)\) equal \(f(-2)\)? No