Question

using the following piecewise functions, find the given values.

g(x) =
\

$$\begin{cases} \\sqrt{5 - x}, & x < -4 \\\\ x^2 - 5, & -4 \\leq x < 2 \\\\ x - 3, & x \\geq 2 \\end{cases}$$

h(x) =
\

$$\begin{cases} -|x|, & x \\leq -5 \\\\ 20 - x^2, & -5 < x \\leq 3 \\\\ 4x - 1, & x > 3 \\end{cases}$$

\lim\limits_{x\to 2^-} g(x) = (2)^2 - 5 = -1
\lim\limits_{x\to -3^+} h(x) =
\lim\limits_{x\to -4^+} g(x) = -(4)^2 - 5 = 11
\lim\limits_{x\to -5} h(x) =
g(2) =
h(3) =
\lim\limits_{x\to -4^-} g(x) =
\lim\limits_{x\to -5^-} h(x) =
\lim\limits_{x\to 2^+} g(x) =
\lim\limits_{x\to 3^+} h(x) =
\lim\limits_{x\to 2} g(x) =
\lim\limits_{x\to 3} h(x) =
\lim\limits_{x\to -4} g(x) =
h(-5) =
g(-4) =
\lim\limits_{x\to 3^-} h(x) =

Answer:

For \( g(x) \)

1. \( \lim_{x \to 2^{-}} g(x) \)

Step1: Determine the relevant piece

For \( x \to 2^{-} \), \( x < 2 \) and \( -4 \leq x < 2 \) (since \( -4 \leq 2^{-} < 2 \)), so use \( g(x)=x^{2}-5 \).

Step2: Substitute \( x = 2 \)

\( \lim_{x \to 2^{-}} (x^{2}-5)=2^{2}-5 = 4 - 5=-1 \)

2. \( \lim_{x \to -4^{+}} g(x) \)

Step1: Determine the relevant piece

For \( x \to -4^{+} \), \( x > - 4 \) and \( -4\leq x < 2 \), so use \( g(x)=x^{2}-5 \).

Step2: Substitute \( x=-4 \)

\( \lim_{x \to -4^{+}} (x^{2}-5)=(-4)^{2}-5=16 - 5 = 11 \)

3. \( g(2) \)

Step1: Determine the relevant piece

For \( x = 2 \), \( x\geq2 \), so use \( g(x)=x - 3 \).

Step2: Substitute \( x = 2 \)

\( g(2)=2-3=-1 \)

4. \( \lim_{x \to -4^{-}} g(x) \)

Step1: Determine the relevant piece

For \( x \to -4^{-} \), \( x < - 4 \), so use \( g(x)=\sqrt{5 - x} \).

Step2: Substitute \( x=-4 \)

\( \lim_{x \to -4^{-}} \sqrt{5 - x}=\sqrt{5-(-4)}=\sqrt{9} = 3 \)

5. \( \lim_{x \to 2^{+}} g(x) \)

Step1: Determine the relevant piece

For \( x \to 2^{+} \), \( x>2 \), so use \( g(x)=x - 3 \).

Step2: Substitute \( x = 2 \)

\( \lim_{x \to 2^{+}} (x - 3)=2-3=-1 \)

6. \( \lim_{x \to 2} g(x) \)

Since \( \lim_{x \to 2^{-}} g(x)=-1 \) and \( \lim_{x \to 2^{+}} g(x)=-1 \), by the definition of limit, \( \lim_{x \to 2} g(x)=-1 \)

7. \( \lim_{x \to -4} g(x) \)

We have \( \lim_{x \to -4^{-}} g(x) = 3 \) and \( \lim_{x \to -4^{+}} g(x)=11 \). Since \( 3
eq11 \), \( \lim_{x \to -4} g(x) \) does not exist.

8. \( g(-4) \)

Step1: Determine the relevant piece

For \( x=-4 \), \( -4\leq x < 2 \), so use \( g(x)=x^{2}-5 \).

Step2: Substitute \( x = - 4 \)

\( g(-4)=(-4)^{2}-5=16 - 5 = 11 \)

For \( h(x) \)

1. \( \lim_{x \to -5^{+}} h(x) \)

Step1: Determine the relevant piece

For \( x \to -5^{+} \), \( x > - 5 \) and \( -5 < x\leq3 \), so use \( h(x)=20 - x^{2} \).

Step2: Substitute \( x=-5 \)

\( \lim_{x \to -5^{+}} (20 - x^{2})=20-(-5)^{2}=20 - 25=-5 \)

2. \( \lim_{x \to -5} h(x) \)

We need to check left - hand limit and right - hand limit.

Step1: Left - hand limit (\( x\to - 5^{-} \))

For \( x\to - 5^{-} \), \( x\leq - 5 \), so use \( h(x)=-|x| \). \( \lim_{x \to -5^{-}} (-|x|)=-|-5|=-5 \)

Step2: Right - hand limit (\( x\to - 5^{+} \))

We already found \( \lim_{x \to -5^{+}} h(x)=-5 \)
Since \( \lim_{x \to -5^{-}} h(x)=\lim_{x \to -5^{+}} h(x)=-5 \), \( \lim_{x \to -5} h(x)=-5 \)

3. \( h(3) \)

Step1: Determine the relevant piece

For \( x = 3 \), \( -5 < x\leq3 \), so use \( h(x)=20 - x^{2} \).

Step2: Substitute \( x = 3 \)

\( h(3)=20-3^{2}=20 - 9 = 11 \)

4. \( \lim_{x \to -5^{-}} h(x) \)

Step1: Determine the relevant piece

For \( x\to - 5^{-} \), \( x\leq - 5 \), so use \( h(x)=-|x| \).

Step2: Substitute \( x=-5 \)

\( \lim_{x \to -5^{-}} (-|x|)=-|-5|=-5 \)

5. \( \lim_{x \to 3^{+}} h(x) \)

Step1: Determine the relevant piece

For \( x \to 3^{+} \), \( x>3 \), so use \( h(x)=4x - 1 \).

Step2: Substitute \( x = 3 \)

\( \lim_{x \to 3^{+}} (4x - 1)=4\times3-1=12 - 1 = 11 \)

6. \( \lim_{x \to 3} h(x) \)

Step1: Left - hand limit (\( x\to 3^{-} \))

For \( x\to 3^{-} \), \( -5 < x\leq3 \), so use \( h(x)=20 - x^{2} \). \( \lim_{x \to 3^{-}} (20 - x^{2})=20-3^{2}=11 \)

Step2: Right - hand limit (\( x\to 3^{+} \))

We already found \( \lim_{x \to 3^{+}} h(x)=11 \)
Since \( \lim_{x \to 3^{-}} h(x)=\lim_{x \to 3^{+}} h(x)=11 \), \( \lim_{x \to 3} h(x)=11 \)

7. \( h(-5) \)

Step1: Determine the relevant piece

For \( x=-5 \), \( x\leq - 5 \), so use \( h(x)=-|x| \).

Step2: Substitute \( x=-5 \)

\( h(-5)=-|-5|=-5 \)

8. \( \lim_{x \to 3^{-}} h(x) \)

Step1: Determine the relevant piece

For \( x\to 3^{-} \), \( -5 < x\leq3 \), so use \( h(x)=20 - x^{2} \).

Step2: Substitute \( x = 3 \)

\( \lim_{x \to 3^{-}} (20 - x^{2})=20-3^{2}=11 \)

Final Answers (for \( g(x) \))

• \( \lim_{x \to 2^{-}} g(x)=\boldsymbol{-1} \)
• \( \lim_{x \to -4^{+}} g(x)=\boldsymbol{11} \)
• \( g(2)=\boldsymbol{-1} \)
• \( \lim_{x \to -4^{-}} g(x)=\boldsymbol{3} \)
• \( \lim_{x \to 2^{+}} g(x)=\boldsymbol{-1} \)
• \( \lim_{x \to 2} g(x)=\boldsymbol{-1} \)
• \( \lim_{x \to -4} g(x) \) does not exist
• \( g(-4)=\boldsymbol{11} \)

Final Answers (for \( h(x) \))

• \( \lim_{x \to -5^{+}} h(x)=\boldsymbol{-5} \)
• \( \lim_{x \to -5} h(x)=\boldsymbol{-5} \)
• \( h(3)=\boldsymbol{11} \)
• \( \lim_{x \to -5^{-}} h(x)=\boldsymbol{-5} \)
• \( \lim_{x \to 3^{+}} h(x)=\boldsymbol{11} \)
• \( \lim_{x \to 3} h(x)=\boldsymbol{11} \)
• \( h(-5)=\boldsymbol{-5} \)
• \( \lim_{x \to 3^{-}} h(x)=\boldsymbol{11} \)