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Removable Discontinuities (When a Limit Exists But the Function is Undefined)

Calculus I

A discontinuity is any point where at least one of the three continuity conditions fails. There are different types, and today's type is called 'removable' — named that because if you simply redefined the function at that one point, the break would disappear, or be 'removed.'

Notes & explanation

What a discontinuity is

A discontinuity is any point where at least one of the three continuity conditions fails. This video covers the removable kind, described as one of the most heavily tested ideas in Unit 1 of AP Calculus.

It is called removable because redefining the function at that single point would make the break disappear.

Formal definition

ff has a removable discontinuity at x=ax = a if lim⁡x→af(x)\displaystyle\lim_{x \to a} f(x) exists but f(a)f(a) is undefined — or, put another way, if

f(a)≠lim⁡x→af(x)f(a) \neq \lim_{x \to a} f(x)

The video makes the point that knowing the formal statement is worth marks on the free-response section on its own, even if the rest of the answer goes wrong.

Worked example

Determine whether f(x)=x2−4x−2f(x) = \dfrac{x^2 - 4}{x - 2} is continuous at x=2x = 2; if not, classify the discontinuity and justify it.

Condition 1 — is f(2)f(2) defined? Direct substitution gives

f(2)=22−42−2=00f(2) = \frac{2^2 - 4}{2 - 2} = \frac{0}{0}

which is undefined, so condition 1 fails immediately and the function is discontinuous at x=2x = 2.

Condition 2 — does the limit still exist? Getting 00\frac{0}{0} is the signal to factor:

f(x)=(x−2)(x+2)(x−2)=x+2,x≠2f(x) = \frac{(x-2)(x+2)}{(x-2)} = x + 2, \quad x \neq 2

Cancelling the (x−2)(x-2) factors is valid precisely because xx is approaching 22 and never equals it — the journey, not the destination. The graph is therefore the line y=x+2y = x + 2 with a hole at x=2x = 2.

lim⁡x→2f(x)=2+2=4\lim_{x \to 2} f(x) = 2 + 2 = 4

So the limit exists and equals 44, while f(2)f(2) does not exist. That is exactly the definition of a removable discontinuity: redefining f(2)=4f(2) = 4 would fill the hole and remove the break.

AP exam tip

Whenever direct substitution gives 00\frac{0}{0}, that is your signal to factor and cancel. The video calls this the single most common limit technique tested on the multiple-choice section.

Common mistake

Do not conclude the limit does not exist just because the function is undefined at that point. Those are two separate conditions in the definition — always simplify and check the limit before deciding.