Question

the graph below is the function $f(x)$
find $limlimits_{x\to -1^-} f(x) = $
find $limlimits_{x\to -1^+} f(x) = $
find $limlimits_{x\to -1} f(x) = $
find $f(-1) = $
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Explanation:

For $\boldsymbol{\lim_{x \to -1^-} f(x)}$:

Step1: Analyze left - hand limit

As $x$ approaches $- 1$ from the left (values of $x$ less than $-1$ and getting closer to $-1$), we look at the part of the graph corresponding to $x < - 1$. The graph is a line segment. We can find the $y$-value that the function approaches as $x$ gets closer to $-1$ from the left.
From the graph, the line for $x < - 1$ passes through points that show that as $x$ approaches $-1$ from the left, the function approaches $1$.
So, $\lim_{x \to -1^-} f(x)=1$.

For $\boldsymbol{\lim_{x \to -1^+} f(x)}$:

Step1: Analyze right - hand limit

As $x$ approaches $-1$ from the right (values of $x$ greater than $-1$ and getting closer to $-1$), we look at the part of the graph corresponding to $x > - 1$. The graph for $x > - 1$ (near $x=-1$) shows that as $x$ approaches $-1$ from the right, the function approaches $-2$.
So, $\lim_{x \to -1^+} f(x)=-2$.

For $\boldsymbol{\lim_{x \to -1} f(x)}$:

Step1: Recall the limit existence condition

The limit $\lim_{x \to a}f(x)$ exists if and only if $\lim_{x \to a^-}f(x)=\lim_{x \to a^+}f(x)$.
We found that $\lim_{x \to -1^-}f(x) = 1$ and $\lim_{x \to -1^+}f(x)=-2$. Since $1
eq - 2$, the two - sided limit $\lim_{x \to -1}f(x)$ does not exist.

For $\boldsymbol{f(-1)}$:

Answer:

s:
$\lim_{x \to -1^-} f(x)=\boldsymbol{1}$

$\lim_{x \to -1^+} f(x)=\boldsymbol{-2}$

$\lim_{x \to -1} f(x)$: Does not exist

$f(-1)=\boldsymbol{4}$