Question

a function $f(x)$ is said to have a jump discontinuity at $x = a$ if:1. $lim_{x \to a^-} f(x)$ exists.2. $lim_{x \to a^+} f(x)$ exists.3. the left and right limits are not equal.let $f(x)=\

$$\begin{cases}4x - 1 & \\text{if } x < 6 \\\\\\frac{2}{x+8} & \\text{if } x \\geq 6\\end{cases}$$

$show that $f(x)$ has a jump discontinuity at $x = 6$ by calculating the limits from the left and right at $x = 6$.$lim_{x \to 6^-} f(x) = \square$$lim_{x \to 6^+} f(x) = \square$now for fun, try to graph $f(x)$.

Explanation:

Step1: Calculate left-hand limit

For $x \to 6^-$, use $f(x)=4x-1$:
$\lim_{x \to 6^-} f(x) = 4(6) - 1$

Step2: Compute left-hand limit value

$\lim_{x \to 6^-} f(x) = 24 - 1 = 23$

Step3: Calculate right-hand limit

For $x \to 6^+$, use $f(x)=\frac{2}{x+8}$:
$\lim_{x \to 6^+} f(x) = \frac{2}{6+8}$

Step4: Compute right-hand limit value

$\lim_{x \to 6^+} f(x) = \frac{2}{14} = \frac{1}{7}$

Step5: Compare the two limits

$23
eq \frac{1}{7}$, so left/right limits exist and are not equal, satisfying jump discontinuity criteria.

Answer:

$\lim_{x \to 6^-} f(x) = 23$
$\lim_{x \to 6^+} f(x) = \frac{1}{7}$
Since the left-hand limit ($23$) and right-hand limit ($\frac{1}{7}$) exist but are not equal, $f(x)$ has a jump discontinuity at $x=6$.