Question

use the function and graph to answer the questions about existence, limits, and continuity.
f(x)=\begin{cases}-x - 3, & -5leq xlt - 4\\x + 5, & -4lt xlt -2\\-\frac{3}{2}x, & -2lt xlt0\\0, & 0leq xleq1\\2x - 2, & 1lt xlt3\\2, & x = 3\\-2x + 10, & 3lt xlt5\\1, & x = 5end{cases}
does f(0) exist?
no
yes
does (lim_{x
ightarrow0}f(x)) exist?
yes
no
does (lim_{x
ightarrow0}f(x)) equal f(0)?
yes
no
is the function continuous at x = 0?
yes
no

Explanation:

Step1: Check f(0) existence

When \(x = 0\), from the piece - wise function \(f(x)=0\) for \(0\leq x\leq1\). So \(f(0) = 0\) exists.

Step2: Check \(\lim_{x

ightarrow0}f(x)\) existence
Find left - hand limit \(\lim_{x
ightarrow0^{-}}f(x)\): For \( - 2ightarrow0^{-}}-\frac{3}{2}x = 0\). Find right - hand limit \(\lim_{x
ightarrow0^{+}}f(x)\): For \(0\leq x\leq1\), \(f(x) = 0\), \(\lim_{x
ightarrow0^{+}}0=0\). Since \(\lim_{x
ightarrow0^{-}}f(x)=\lim_{x
ightarrow0^{+}}f(x)=0\), \(\lim_{x
ightarrow0}f(x)\) exists.

Step3: Compare \(\lim_{x

ightarrow0}f(x)\) and \(f(0)\)
We have \(\lim_{x
ightarrow0}f(x) = 0\) and \(f(0)=0\), so \(\lim_{x
ightarrow0}f(x)=f(0)\).

Step4: Check continuity at \(x = 0\)

A function \(y = f(x)\) is continuous at \(x=a\) if \(\lim_{x
ightarrow a}f(x)=f(a)\). Since \(\lim_{x
ightarrow0}f(x)=f(0) = 0\), the function is continuous at \(x = 0\).

Answer:

Does \(f(0)\) exist? Yes
Does \(\lim_{x
ightarrow0}f(x)\) exist? Yes
Does \(\lim_{x
ightarrow0}f(x)\) equal \(f(0)\)? Yes
Is the function continuous at \(x = 0\)? Yes