Question

Please help verify if my answers are correct! Thank you:)

Answer 1

From the graph of the function giving, Consider the image below

`\begin{gathered} \text{when x=-1.f(-1)=1} \\ \text{Hence } \\ f(-1)=1 \end{gathered}`

Hence

F(-1)=1

From the graph,

hence

F(1)=1

For

`\begin{gathered} \text{when x=6,f(6)=0} \\ \text{Hence } \\ f(6)=0 \end{gathered}`

Hence

f(6)=0

The limit of f(x) as x approaches zero

`\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}`

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

`\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}`

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

`\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}`

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