Question

2.2 the limit of a function: problem 8 (6 points) sketch the graph of the following function and use it to determine the following limits. if a limit does not exist, type dne.

$$ f(x)=\
LATEXBLOCK0
$$

1.

$$ \\lim _{x \ ightarrow -1^{-}} f(x)= $$

2.

$$ \\lim _{x \ ightarrow -1^{+}} f(x)= $$

3.

$$ \\lim _{x \ ightarrow -1} f(x)= $$

4.

$$ \\lim _{x \ ightarrow 1} f(x)= $$

1. $$ f(-1)= $$

Explanation:

Step1: Analyze the left - hand limit as \(x\to - 1^{-}\)

For \(x\to - 1^{-}\), we use the function \(f(x)=x - 5\). Substitute \(x=-1\) into \(f(x)=x - 5\), we get \(f(-1)=-1 - 5=-6\). So \(\lim_{x\to - 1^{-}}f(x)=-6\).

Step2: Analyze the right - hand limit as \(x\to - 1^{+}\)

For \(x\to - 1^{+}\), we use the function \(f(x)=x^{2}+4\). Substitute \(x = - 1\) into \(f(x)=x^{2}+4\), we get \(f(-1)=(-1)^{2}+4=1 + 4=1\). So \(\lim_{x\to - 1^{+}}f(x)=1\).

Step3: Determine \(\lim_{x\to - 1}f(x)\)

Since \(\lim_{x\to - 1^{-}}f(x)
eq\lim_{x\to - 1^{+}}f(x)\) (i.e., \(-6
eq1\)), by the definition of the limit (the limit of a function \(y = f(x)\) as \(x\to a\) exists if and only if \(\lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)\)), \(\lim_{x\to - 1}f(x)\) does not exist.

Step4: Analyze \(\lim_{x\to1}f(x)\)

For \(x\to1\), we need to check the left - hand limit (\(x\to1^{-}\)) and the right - hand limit (\(x\to1^{+}\)).

• For \(x\to1^{-}\), use \(f(x)=x^{2}+4\). Substitute \(x = 1\) into \(f(x)=x^{2}+4\), we get \(f(1)=1^{2}+4=5\).
• For \(x\to1^{+}\), use \(f(x)=6 - x\). Substitute \(x = 1\) into \(f(x)=6 - x\), we get \(f(1)=6-1 = 5\).

Since \(\lim_{x\to1^{-}}f(x)=\lim_{x\to1^{+}}f(x)=5\), by the definition of the limit, \(\lim_{x\to1}f(x)=5\).

Step5: Find \(f(-1)\)

Since \(x=-1\) satisfies \(x\leq - 1\), we use \(f(x)=x - 5\). Substitute \(x=-1\) into \(f(x)=x - 5\), we get \(f(-1)=-1-5=-6\).

Answer:

1. \(-6\)
2. \(1\)
3. DNE
4. \(5\)
5. \(-6\)