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Math · Calculus I · Worked example

Classify three kinds of discontinuity

Classify the discontinuity of each function: f(x) = (x² − 9)/(x − 3) at x = 3, g(x) = |x|/x at x = 0 and h(x) = 1/(x − 3)² at x = 3.

A removable discontinuity

f(3) is undefined, but for x ≠ 3, f(x) = x + 3, which approaches 6. The limit exists, so the hole at (3, 6) is removable: defining f(3) = 6 would fill it.

limx→3x2−9x−3=limx→3(x+3)=6
limx→3x2−9x−3=limx→3(x+3)=6

A jump discontinuity

|x|/x is −1 for x < 0 and 1 for x > 0. Both one-sided limits exist, but they differ, so the graph jumps.

limx→0−∣⁢x⁢∣x=−1,limx→0+∣⁢x⁢∣x=1
limx→0−∣⁢x⁢∣x=−1limx→0+∣⁢x⁢∣x=1

An infinite discontinuity

As x approaches 3 from either side, (x − 3)² is a small positive number, so h(x) grows without bound: the line x = 3 is a vertical asymptote.

limx→31(x−3)2=∞

Result

f is removable (a hole at (3, 6)), g jumps from −1 to 1, and h is infinite (a vertical asymptote at x = 3).

Your turn

Classify the discontinuity of k(x) = (x² − 1)/(x + 1) at x = −1.

Show the answer and explanation

Removable: the limit is −2, but k(−1) is undefined.

For x ≠ −1, k(x) = x − 1, which approaches −2.

limx→−1(x−1)=−2

Keep exploring

In Limits & one-sided behavior, |x|/x at 0 shows a left limit of −1 and a right limit of 1, so the limit does not exist. Change the function to (x² − 9)/(x − 3) and the approach to 3: both sides give 6.

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