Solved on Jan 10, 2024

Determine if the function $f(x)$ has a jump discontinuity at $x=8$ by finding $\lim _{x \rightarrow 8^{-}} f(x)$ and $\lim _{x \rightarrow 8^{+}} f(x)$ .

STEP 1

Assumptions
1. The function $f(x)$ is defined piecewise with two different expressions for $x<8$ and $x\geq8$ .
2. To show a jump discontinuity at $x=8$ , we need to calculate the left-hand limit $\lim _{x \rightarrow 8^{-}} f(x)$ and the right-hand limit $\lim _{x \rightarrow 8^{+}} f(x)$ .
3. A jump discontinuity occurs if both limits exist but are not equal.

STEP 2

Calculate the left-hand limit $\lim _{x \rightarrow 8^{-}} f(x)$ . For $x<8$ , the function $f(x)$ is defined as $5x-5$ .
$\lim _{x \rightarrow 8^{-}} f(x) = \lim _{x \rightarrow 8^{-}} (5x-5)$

STEP 3

Evaluate the left-hand limit by substituting $x$ with $8$ directly, since the function is continuous for $x<8$ .
$\lim _{x \rightarrow 8^{-}} (5x-5) = 5 \cdot 8 - 5$

STEP 4

Calculate the result of the left-hand limit.
$\lim _{x \rightarrow 8^{-}} (5x-5) = 40 - 5 = 35$

STEP 5

Now, calculate the right-hand limit $\lim _{x \rightarrow 8^{+}} f(x)$ . For $x\geq8$ , the function $f(x)$ is defined as $\frac{4}{x+9}$ .
$\lim _{x \rightarrow 8^{+}} f(x) = \lim _{x \rightarrow 8^{+}} \left(\frac{4}{x+9}\right)$

STEP 6

Evaluate the right-hand limit by substituting $x$ with $8$ directly, since the function is continuous for $x\geq8$ .
$\lim _{x \rightarrow 8^{+}} \left(\frac{4}{x+9}\right) = \frac{4}{8+9}$

STEP 7

Calculate the result of the right-hand limit.
$\lim _{x \rightarrow 8^{+}} \left(\frac{4}{x+9}\right) = \frac{4}{17}$

STEP 8

Compare the left-hand limit and the right-hand limit to determine if there is a jump discontinuity at $x=8$ .
$\lim _{x \rightarrow 8^{-}} f(x) = 35 \neq \frac{4}{17} = \lim _{x \rightarrow 8^{+}} f(x)$

STEP 9

Since the left-hand limit and the right-hand limit are not equal, we conclude that $f(x)$ has a jump discontinuity at $x=8$ .

STEP 10

For fun, let's graph the function $f(x)$ . To graph $f(x)$ , we need to plot the two pieces of the function separately and indicate the discontinuity at $x=8$ .
1. For $x<8$ , graph the line $y=5x-5$ .
2. For $x\geq8$ , graph the curve $y=\frac{4}{x+9}$ .
3. Mark the point where $x=8$ and show that there is a jump from the value of the left-hand limit to the value of the right-hand limit.
The graph would show a straight line for $x<8$ ending at the point $(8,35)$ , and for $x\geq8$ , it would start from a point just above the $x$ -axis at $(8,\frac{4}{17})$ and approach the $x$ -axis as $x$ increases.

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Determine if the function $f(x)$ has a jump discontinuity at $x=8$ by finding $\lim _{x \rightarrow 8^{-}} f(x)$ and $\lim _{x \rightarrow 8^{+}} f(x)$ .

STEP 1

STEP 2

Calculate the left-hand limit $\lim _{x \rightarrow 8^{-}} f(x)$ . For $x<8$ , the function $f(x)$ is defined as $5x-5$ .
$\lim _{x \rightarrow 8^{-}} f(x) = \lim _{x \rightarrow 8^{-}} (5x-5)$

STEP 3

Evaluate the left-hand limit by substituting $x$ with $8$ directly, since the function is continuous for $x<8$ .
$\lim _{x \rightarrow 8^{-}} (5x-5) = 5 \cdot 8 - 5$

STEP 4

Calculate the result of the left-hand limit.
$\lim _{x \rightarrow 8^{-}} (5x-5) = 40 - 5 = 35$

STEP 5

Now, calculate the right-hand limit $\lim _{x \rightarrow 8^{+}} f(x)$ . For $x\geq8$ , the function $f(x)$ is defined as $\frac{4}{x+9}$ .
$\lim _{x \rightarrow 8^{+}} f(x) = \lim _{x \rightarrow 8^{+}} \left(\frac{4}{x+9}\right)$

STEP 6

Evaluate the right-hand limit by substituting $x$ with $8$ directly, since the function is continuous for $x\geq8$ .
$\lim _{x \rightarrow 8^{+}} \left(\frac{4}{x+9}\right) = \frac{4}{8+9}$

STEP 7

Calculate the result of the right-hand limit.
$\lim _{x \rightarrow 8^{+}} \left(\frac{4}{x+9}\right) = \frac{4}{17}$

STEP 8

Compare the left-hand limit and the right-hand limit to determine if there is a jump discontinuity at $x=8$ .
$\lim _{x \rightarrow 8^{-}} f(x) = 35 \neq \frac{4}{17} = \lim _{x \rightarrow 8^{+}} f(x)$

STEP 9

Since the left-hand limit and the right-hand limit are not equal, we conclude that $f(x)$ has a jump discontinuity at $x=8$ .

STEP 10

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Determine if the function f(x)f(x)f(x) has a jump discontinuity at x=8x=8x=8 by finding lim⁡x→8−f(x)\lim _{x \rightarrow 8^{-}} f(x)limx→8−​f(x) and lim⁡x→8+f(x)\lim _{x \rightarrow 8^{+}} f(x)limx→8+​f(x).

STEP 1

STEP 2

STEP 3

Evaluate the left-hand limit by substituting xxx with 888 directly, since the function is continuous for x<8x<8x<8.lim⁡x→8−(5x−5)=5⋅8−5\lim _{x \rightarrow 8^{-}} (5x-5) = 5 \cdot 8 - 5x→8−lim​(5x−5)=5⋅8−5

STEP 4

Calculate the result of the left-hand limit.lim⁡x→8−(5x−5)=40−5=35\lim _{x \rightarrow 8^{-}} (5x-5) = 40 - 5 = 35x→8−lim​(5x−5)=40−5=35

STEP 5

STEP 6

Evaluate the right-hand limit by substituting xxx with 888 directly, since the function is continuous for x≥8x\geq8x≥8.lim⁡x→8+(4x+9)=48+9\lim _{x \rightarrow 8^{+}} \left(\frac{4}{x+9}\right) = \frac{4}{8+9}x→8+lim​(x+94​)=8+94​

STEP 7

Calculate the result of the right-hand limit.lim⁡x→8+(4x+9)=417\lim _{x \rightarrow 8^{+}} \left(\frac{4}{x+9}\right) = \frac{4}{17}x→8+lim​(x+94​)=174​

STEP 8

Compare the left-hand limit and the right-hand limit to determine if there is a jump discontinuity at x=8x=8x=8.lim⁡x→8−f(x)=35≠417=lim⁡x→8+f(x)\lim _{x \rightarrow 8^{-}} f(x) = 35 \neq \frac{4}{17} = \lim _{x \rightarrow 8^{+}} f(x)x→8−lim​f(x)=35=174​=x→8+lim​f(x)

STEP 9

Since the left-hand limit and the right-hand limit are not equal, we conclude that f(x)f(x)f(x) has a jump discontinuity at x=8x=8x=8.

STEP 10

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Determine if the function $f(x)$ has a jump discontinuity at $x=8$ by finding $\lim _{x \rightarrow 8^{-}} f(x)$ and $\lim _{x \rightarrow 8^{+}} f(x)$ .

Evaluate the left-hand limit by substituting $x$ with $8$ directly, since the function is continuous for $x<8$ .
$\lim _{x \rightarrow 8^{-}} (5x-5) = 5 \cdot 8 - 5$

Calculate the result of the left-hand limit.
$\lim _{x \rightarrow 8^{-}} (5x-5) = 40 - 5 = 35$

Evaluate the right-hand limit by substituting $x$ with $8$ directly, since the function is continuous for $x\geq8$ .
$\lim _{x \rightarrow 8^{+}} \left(\frac{4}{x+9}\right) = \frac{4}{8+9}$

Calculate the result of the right-hand limit.
$\lim _{x \rightarrow 8^{+}} \left(\frac{4}{x+9}\right) = \frac{4}{17}$

Compare the left-hand limit and the right-hand limit to determine if there is a jump discontinuity at $x=8$ .
$\lim _{x \rightarrow 8^{-}} f(x) = 35 \neq \frac{4}{17} = \lim _{x \rightarrow 8^{+}} f(x)$

Since the left-hand limit and the right-hand limit are not equal, we conclude that $f(x)$ has a jump discontinuity at $x=8$ .