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SOLUTION
From the graph of the function giving, Consider the image below
[tex]\begin{gathered} \text{when x=-1.f(-1)=1} \\ \text{Hence } \\ f(-1)=1 \end{gathered}[/tex]
Hence
F(-1)=1
From the graph,
[tex]\begin{gathered} \text{when }x=1,f(1) \\ \text{Hence } \\ f(1)=1 \\ \text{The unshaded point shows the number is not included } \end{gathered}[/tex]hence
F(1)=1
For
[tex]\begin{gathered} \text{when x=6,f(6)=0} \\ \text{Hence } \\ f(6)=0 \end{gathered}[/tex]Hence
f(6)=0
The limit of f(x) as x approaches zero
[tex]\begin{gathered} \text{The Right hand limit is } \\ \lim _{x\rightarrow0^+}f(x)=-1 \\ \text{and } \\ \text{The left hand limit} \\ \lim _{x\rightarrow0^-}f(x)=0 \end{gathered}[/tex]Since the Left hand limit is not equal to the Right Linit (DLR)
The limit does not exist(DNE)
The limit of f(x) as x approaches two is
[tex]\begin{gathered} \text{The right hand limit is } \\ \lim _{x\rightarrow2^+}f(x)=0 \\ \text{And } \\ \text{The left hand limit is } \\ \lim _{x\rightarrow2^-}f(x)=0 \end{gathered}[/tex]The Left hand limit equalis the right hand Limit,
Hence Limit has x approaches 2 is 0
The limit of f(x) as x approaches 4 is given as
[tex]\begin{gathered} \text{The right hand limit is } \\ \lim _{x\rightarrow4^+}f(x)=-2 \\ \text{And The right hand limit is } \\ \lim _{x\rightarrow4^-}f(x)=-2 \end{gathered}[/tex]From the graph, the function is not continous at f(x)=-2, but the limit exist at that point since the Left hand limit and the right hand limit are the same.
Hence
Limit as x approaches 4 is -2