You remove a discontinuity by finding the limit of the function at the problem point and then redefining the function’s value at that point to equal the limit. This only works for removable discontinuities, the type that appears as a small hole in an otherwise smooth graph. Jump discontinuities and vertical asymptotes cannot be fixed this way.
What Makes a Discontinuity Removable
A function is continuous at a point when three conditions are all true: the function is defined at that point, the limit exists as you approach that point, and the function’s value actually equals the limit. A removable discontinuity breaks the third condition (or sometimes the first). The limit exists and is finite, but the function either isn’t defined there or is defined as the wrong value. That mismatch is the entire problem, and it’s fixable.
On a graph, a removable discontinuity looks like a small open circle, a single missing or mismatched point on an otherwise smooth curve. Think of it as a hole in a floor that you can patch. By contrast, a jump discontinuity is like a step down where two pieces of the graph end at different heights, and an infinite discontinuity is a vertical asymptote where the function shoots off toward infinity. Neither of those has a single finite limit you can plug in, so neither can be “removed” by redefining one point.
The Core Idea in Three Steps
The process is the same regardless of the function type:
- Step 1: Find the trouble spot. Identify the x-value where the function is undefined or where its value doesn’t match the surrounding behavior.
- Step 2: Calculate the limit. Determine what value the function approaches as x gets closer and closer to that point from both sides. If this two-sided limit exists and is a finite number L, the discontinuity is removable.
- Step 3: Redefine the function. Create a new piecewise version of the function that equals the original everywhere except at the problem point, where you assign it the value L.
That’s it. You now have a function that is continuous at that point because all three continuity conditions are satisfied: it’s defined, the limit exists, and the value equals the limit.
Removing a Hole in a Rational Function
The most common scenario you’ll encounter is a rational function (a fraction with polynomials on top and bottom) where plugging in a specific x-value gives you 0/0. This indeterminate form is a strong signal that a removable discontinuity exists.
Take the function f(x) = (x² – 4) / (x – 2). Plugging in x = 2 gives 0/0, so the function is undefined there. But you can factor the numerator:
(x² – 4) / (x – 2) = (x – 2)(x + 2) / (x – 2)
Cancel the common factor (x – 2) to get (x + 2). Now evaluate the simplified expression at x = 2: you get 4. The limit as x approaches 2 is 4, so you remove the discontinuity by defining f(2) = 4. The graph of this function is just the line y = x + 2 with a tiny hole at the point (2, 4). Your redefinition fills that hole.
Here’s a second example with a slightly less obvious factoring. Consider f(x) = (2x – 6) / (x² + x – 12). Both the numerator and denominator equal zero at x = 3. Factor them:
2(x – 3) / [(x + 4)(x – 3)]
Cancel the (x – 3) terms, leaving 2 / (x + 4). Evaluate at x = 3 to get 2/7. To remove the discontinuity, define f(3) = 2/7.
Writing the Piecewise Definition
Once you’ve found the limit, you formally express the repaired function as a piecewise function. For the first example above, you’d write:
F(x) = { (x² – 4)/(x – 2) when x ≠ 2, and 4 when x = 2 }
This new function F is identical to the original f at every single point except x = 2, where it now has the correct value. It is continuous everywhere the original was continuous, plus at x = 2. In many textbook problems, you’ll be asked to “find the value of k that makes the function continuous,” where k is the function’s value at the discontinuity. The answer is always the limit.
When a Discontinuity Cannot Be Removed
Not every discontinuity cooperates. If the left-side limit and right-side limit approach different values, you have a jump discontinuity. No single number can bridge that gap. If the function heads toward positive or negative infinity near the point, you have an infinite discontinuity (a vertical asymptote), and no finite value can fix that either.
A quick diagnostic: if direct substitution gives you 0/0, try factoring or simplifying. The discontinuity is likely removable. If substitution gives you something like 5/0, the function blows up and the discontinuity is not removable. And if the function is defined by separate rules on each side of a point and those rules approach different values, that’s a jump, also not removable.
Techniques Beyond Simple Factoring
Factoring and canceling works for most rational functions, but some problems require other limit-finding techniques to identify the value you need.
For functions involving square roots, multiply by the conjugate. For trigonometric functions, you may need standard limit identities like the fact that sin(x)/x approaches 1 as x approaches 0. For more complex expressions, L’Hôpital’s rule can resolve 0/0 forms by taking the derivative of the top and bottom separately and then evaluating.
Regardless of the technique, the underlying logic never changes. You’re finding the limit at the problem point, then assigning that value to the function. The method for computing the limit varies; the method for removing the discontinuity does not.