Question

a function f(x) is said to have a jump discontinuity at x = a if: 1. lim_{x→a^-} f(x) exists. 2. lim_{x→a^+} f(x) exists. 3. the left and right limits are not equal. let f(x) = {4x - 7, if x < 3; 4/(x + 6), if x ≥ 3}. show that f(x) has a jump discontinuity at x = 3 by calculating the limits from the left and right at x = 3. lim_{x→3^-} f(x) = □ lim_{x→3^+} f(x) = □ now, for fun, try to graph f(x).

Explanation:

Step1: Calculate left - hand limit

For $x\to3^{-}$, we use the part of the function $f(x)=4x - 7$ since $x<3$.
$\lim_{x\to3^{-}}f(x)=\lim_{x\to3^{-}}(4x - 7)=4\times3-7=12 - 7 = 5$.

Step2: Calculate right - hand limit

For $x\to3^{+}$, we use the part of the function $f(x)=\frac{4}{x + 6}$ since $x\geq3$.
$\lim_{x\to3^{+}}f(x)=\lim_{x\to3^{+}}\frac{4}{x + 6}=\frac{4}{3+6}=\frac{4}{9}$.

Answer:

$\lim_{x\to3^{-}}f(x)=5$, $\lim_{x\to3^{+}}f(x)=\frac{4}{9}$