Question

1. use the graph to find the limit (if it exist). if the limit does not exit, explain why

y = f(x)
lim_{x
ightarrow - 2^{+}}f(x)= lim_{x
ightarrow - 2^{-}}f(x)= lim_{x
ightarrow - 2}f(x)=f(-2)= lim_{x
ightarrow - 1^{+}}f(x)= lim_{x
ightarrow - 1^{-}}f(x)= lim_{x
ightarrow - 1}f(x)=f(-1)= lim_{x
ightarrow 2^{+}}f(x)= lim_{x
ightarrow 2^{-}}f(x)= lim_{x
ightarrow 2}f(x)=f(2)= lim_{x
ightarrow 0^{+}}f(x)= lim_{x
ightarrow 0^{-}}f(x)= lim_{x
ightarrow 0}f(x)=f(0)=

Explanation:

Step1: Analyze right - hand limit as $x\to - 2^{+}$

As $x$ approaches $-2$ from the right, we look at the part of the graph for $x > - 2$ near $x=-2$. The value of $y = f(x)$ approaches $0$. So, $\lim_{x\to - 2^{+}}f(x)=0$.

Step2: Analyze left - hand limit as $x\to - 2^{-}$

As $x$ approaches $-2$ from the left, we look at the part of the graph for $x < - 2$ near $x = - 2$. The value of $y=f(x)$ approaches $0$. So, $\lim_{x\to - 2^{-}}f(x)=0$.

Step3: Determine the limit as $x\to - 2$

Since $\lim_{x\to - 2^{+}}f(x)=\lim_{x\to - 2^{-}}f(x)=0$, then $\lim_{x\to - 2}f(x)=0$. And from the graph, $f(-2)$ is undefined (there is a hole at $x = - 2$).

Step4: Analyze right - hand limit as $x\to - 1^{+}$

As $x$ approaches $-1$ from the right, the value of $y = f(x)$ approaches $1$. So, $\lim_{x\to - 1^{+}}f(x)=1$.

Step5: Analyze left - hand limit as $x\to - 1^{-}$

As $x$ approaches $-1$ from the left, the value of $y = f(x)$ approaches $1$. So, $\lim_{x\to - 1^{-}}f(x)=1$.

Step6: Determine the limit as $x\to - 1$

Since $\lim_{x\to - 1^{+}}f(x)=\lim_{x\to - 1^{-}}f(x)=1$, then $\lim_{x\to - 1}f(x)=1$. And $f(-1)=1$.

Step7: Analyze right - hand limit as $x\to 2^{+}$

As $x$ approaches $2$ from the right, the value of $y = f(x)$ approaches $- 2$. So, $\lim_{x\to 2^{+}}f(x)=-2$.

Step8: Analyze left - hand limit as $x\to 2^{-}$

As $x$ approaches $2$ from the left, the value of $y = f(x)$ approaches $-2$. So, $\lim_{x\to 2^{-}}f(x)=-2$.

Step9: Determine the limit as $x\to 2$

Since $\lim_{x\to 2^{+}}f(x)=\lim_{x\to 2^{-}}f(x)=-2$, then $\lim_{x\to 2}f(x)=-2$. And $f(2)$ is undefined (there is a hole at $x = 2$).

Step10: Analyze right - hand limit as $x\to 0^{+}$

As $x$ approaches $0$ from the right, the value of $y = f(x)$ approaches $-1$. So, $\lim_{x\to 0^{+}}f(x)=-1$.

Step11: Analyze left - hand limit as $x\to 0^{-}$

As $x$ approaches $0$ from the left, the value of $y = f(x)$ approaches $2$. So, $\lim_{x\to 0^{-}}f(x)=2$.

Step12: Determine the limit as $x\to 0$

Since $\lim_{x\to 0^{+}}f(x)
eq\lim_{x\to 0^{-}}f(x)$, $\lim_{x\to 0}f(x)$ does not exist. And $f(0)=2$.

Answer:

$\lim_{x\to - 2^{+}}f(x)=0$
$\lim_{x\to - 2^{-}}f(x)=0$
$\lim_{x\to - 2}f(x)=0$
$f(-2)$ is undefined
$\lim_{x\to - 1^{+}}f(x)=1$
$\lim_{x\to - 1^{-}}f(x)=1$
$\lim_{x\to - 1}f(x)=1$
$f(-1)=1$
$\lim_{x\to 2^{+}}f(x)=-2$
$\lim_{x\to 2^{-}}f(x)=-2$
$\lim_{x\to 2}f(x)=-2$
$f(2)$ is undefined
$\lim_{x\to 0^{+}}f(x)=-1$
$\lim_{x\to 0^{-}}f(x)=2$
$\lim_{x\to 0}f(x)$ does not exist
$f(0)=2$