Question

a function $f(x)$ is said to have a jump discontinuity at $x = a$ if:

1. $lim_{x

ightarrow a^{-}}f(x)$ exists.

1. $lim_{x

ightarrow a^{+}}f(x)$ exists.

1. the left and right limits are not equal.

let $f(x)=\begin{cases}7x - 1&\text{if }x < 10\\frac{5}{x + 5}&\text{if }xgeq10end{cases}$
show that $f(x)$ has a jump discontinuity at $x = 10$ by calculating the limits from the left and right at $x = 10$.
$lim_{x
ightarrow10^{-}}f(x)=$
$lim_{x
ightarrow10^{+}}f(x)=$
now for fun, try to graph $f(x)$.
question help: message instructor

Explanation:

Step1: Calculate left - hand limit

For $x\to10^{-}$, we use the part of the function $f(x)=7x - 1$ since $x<10$.
$\lim_{x\to10^{-}}f(x)=\lim_{x\to10^{-}}(7x - 1)=7\times10-1=69$

Step2: Calculate right - hand limit

For $x\to10^{+}$, we use the part of the function $f(x)=\frac{5}{x + 5}$ since $x\geq10$.
$\lim_{x\to10^{+}}f(x)=\lim_{x\to10^{+}}\frac{5}{x + 5}=\frac{5}{10 + 5}=\frac{5}{15}=\frac{1}{3}$

Answer:

$\lim_{x\to10^{-}}f(x)=69$
$\lim_{x\to10^{+}}f(x)=\frac{1}{3}$