Question

Let f(x)= 9&if x<-4\\ -x+5&if-4<= x<4\\ -2&if x=4\\ 5& ifx >4.

Sketch the graph of this function and find the following limits, if they exist. (If a limit does not exist, enter DNE.) lim f(x)=
1. --4- lim f(x)=
2. →−4+ lim f(x)=
3. -4 lim f(x)=
4. lim f(x)=
5. x+4+ lim f(x)=
6. x+4+ lim f(x)=

Answer 1

`\sf\:f(x) = \begin{cases}9 & \text{if } x < -4 \\ -x+5 & \text{if } -4 \leq x < 4 \\ -2 & \text{if } x = 4 \\ 5 & \text{if } x > 4 \\ \end{cases} \\`

To sketch the graph of this function, we plot the points and lines as follows:

`\sf\:\begin{align}(-\infty, -4) & : \text{Line segment with a constant value of } 9 \\ [-4, 4) & : \text{Line segment with a slope of -1 and y-intercept of 5} \\ (4, \infty) & : \text{Horizontal line with a constant value of } 5 \\ x = 4 & : \text{Point at } (4, -2) \\ \end{align} \\`

1.
`\sf\:\lim_{{x \to -4^-}} f(x) \\`: The limit as x approaches -4 from the left side. Since the function is continuous at -4, the limit exists and is equal to the value of the function at that point. So,
`\sf\:\lim_{{x \to -4^-}} f(x) = f(-4) = 9 \\`.

2.
`\sf\:\lim_{{x \to -4^+}} f(x) \\`: The limit as x approaches -4 from the right side. Again, since the function is continuous at -4 , the limit exists and is equal to the value of the function at that point. So,
`\sf\:\lim_{{x \to -4^+}} f(x) = f(-4) = 9 \\`.

3.
`\sf\:\lim_{{x \to -4}} f(x) \\`: The limit as x approaches -4. Since the left and right limits both exist and are equal, the overall limit exists and is equal to the common value. So,
`\sf\:\lim_{{x \to -4}} f(x) = \lim_{{x \to -4^-}} f(x) = \lim_{{x \to -4^+}} f(x) = 9 \\`.

4.
`\sf\:\lim_{{x \to 4}} f(x) \\`: The limit as x approaches 4. Since the function has a discontinuity at
`\sf\:x = 4 \\` (a jump from
`\sf\:-x + 5 \\` to (-2), the limit does not exist. So,
`\sf\:\lim_{{x \to 4}} f(x) \\` is DNE.

5.
`\sf\:\lim_{{x \to 4^+}} f(x) \\`: The limit as x approaches 4 from the right side. Since the function is continuous at 4, the limit exists and is equal to the value of the function at that point. So,
`\sf\:\lim_{{x \to 4^+}} f(x) = f(4) = -2 \\`.

6.
`\sf\:\lim_{{x \to 4^+}} (x + 4) f(x) \\`: The limit as x approaches 4 from the right side, multiplied by
`\sf\:(x + 4) \\`. Since the function is continuous at 4, we can evaluate this limit by substituting

`\sf\:x = 4. So, \lim_{{x \to 4^+}} (x + 4) f(x) = (4 + 4) f(4) = 8 \cdot (-2) = -16 \\`.

That's it!