How to Tell If a Function Is Continuous or Discontinuous: 3 Tests

A function is continuous at a point x = c when three things hold at once: f© is defined, the limit of f(x) as x approaches c exists, and that limit equals f©. If any one of those fails, the function is discontinuous there. Knowing how to tell if a function is continuous or discontinuous comes down to running three quick tests: check the three conditions, inspect the graph for breaks, or evaluate the limits algebraically.

What Continuity Means

Continuity at a single point c means the curve passes through (c, f©) without a break. The precise definition: f is continuous at c if lim(x→c) f(x) = f©. Everything hinges on that equality.

Continuity on an interval is just pointwise continuity everywhere inside it. A function continuous on the open interval (a, b) has no interruptions between a and b. On a closed interval [a, b], we also require one-sided continuity at the endpoints: the limit from the right at a equals f(a), and the limit from the left at b equals f(b).

Continuous behavior means values change smoothly as x moves. Discontinuous behavior means at some x-value the graph breaks — a hole, a jump, or a blow-up toward infinity. These notes from Paul’s Online Math Notes lay out the same three-part definition most calculus courses use.

Test 1: Check the Three Conditions

The formal test is a checklist. At the candidate point x = c, verify each item in order.

1. f© exists. The point c must be in the domain, so plugging in c gives a real number. If f© is undefined, the function is discontinuous at c — stop there.
2. The limit exists. As x approaches c from both sides, the values must settle on one finite number L. If the left side and right side head to different values, the limit does not exist.
3. The limit equals the value. Confirm lim(x→c) f(x) = f©. This catches graphs that “look” fine but have a point plotted in the wrong place.

A quick worked example

Let f(x) = (x² − 1)/(x − 1) for x ≠ 1, and f(1) = 3.

• Step 1: f(1) = 3, so the value exists.
• Step 2: factoring gives (x − 1)(x + 1)/(x − 1) = x + 1 for x ≠ 1, so lim(x→1) f(x) = 2.
• Step 3: the limit (2) does not equal f(1) (3).

The function is discontinuous at x = 1 because the third condition fails. If instead f(1) had been defined as 2, it would be continuous. Khan Academy has practice sets on exactly this three-step check.

Test 2: Spot Discontinuities Graphically

A graph makes discontinuities visible if you know the three shapes to look for.

• Holes (removable discontinuities): a single missing point in an otherwise smooth curve. The limit exists, but f© is undefined or mismatched. The function above has a hole at x = 1.
• Jumps (jump discontinuities): the curve stops and restarts at a different height. The two one-sided limits exist but disagree, so the limit does not exist. Step functions and piecewise functions are common sources.
• Asymptotes (infinite discontinuities): the graph shoots toward ±∞ near a vertical line, as with f(x) = 1/x at x = 0. The limit is unbounded, so it does not exist.

Removable vs. non-removable

The distinction matters because one can be patched and the other cannot. A removable discontinuity is a hole — you could redefine the single point f© to match the limit and make the function continuous there. Jumps and asymptotes are non-removable: no single value at c can bridge a jump or tame an infinite spike.

When reading a graph, trace it left to right with a pencil. Anywhere you must lift the pencil is a discontinuity, and the way you lift it tells you which type.

Test 3: Evaluate Limits Algebraically

When no graph is available, compute the limit directly. The key is comparing the one-sided limits, because a two-sided limit exists only when both sides agree.

• Direct substitution: if plugging c into f gives a finite number, the limit equals f© and you are done in one step. Polynomials, for example, are continuous everywhere.
• Algebraic repair: a 0/0 result signals a hole. Factor, cancel, or rationalize, then substitute again — this is how the example above produced the limit 2.
• One-sided limits: evaluate lim(x→c⁻) and lim(x→c⁺) separately. If both give the same finite L, then lim(x→c) f(x) = L. If they differ, the two-sided limit does not exist.

One-sided limits in action

Take the step function g(x) = 1 for x < 0 and g(x) = 2 for x ≥ 0. From the left, lim(x→0⁻) g(x) = 1. From the right, lim(x→0⁺) g(x) = 2. Since 1 ≠ 2, the overall limit does not exist, so g is discontinuous at x = 0 — a jump. Compare that with f(x) = |x|/x, where the sides give −1 and +1; same verdict, different shape.

Once the limit is known, close the loop from Test 1: does it equal f©? If yes, continuous at c; if no or undefined, discontinuous. MIT OpenCourseWare covers limit techniques in depth, including the one-sided comparisons used here. For piecewise functions, always test the boundary points where the rule changes — that is where discontinuities almost always hide.

Piecewise Functions: Where Continuity Usually Breaks

Piecewise functions are a common setting for discontinuity questions, because the rule changes create natural fault lines. The procedure is mechanical: test each boundary point separately, then test the interior of each piece.

Consider h(x) = x + 2 for x < 1, and h(x) = x² + 2 for x ≥ 1. Inside each piece the function is a polynomial, hence continuous — no work needed there. The only possible break is at x = 1.

• Value: h(1) = 1² + 2 = 3, taken from the second piece since the condition is x ≥ 1.
• Left limit: lim(x→1⁻) (x + 2) = 3.
• Right limit: lim(x→1⁺) (x² + 2) = 3.

Both sides agree and equal h(1), so h is continuous at x = 1 and therefore everywhere.

Now change the second piece to x² + 5. The left limit stays 3, the right limit becomes 6, and h(1) = 6. The one-sided limits disagree, so the limit fails to exist — a jump discontinuity at x = 1, non-removable because no single point value can satisfy both sides.

Common Traps

A few situations reliably generate wrong answers.

Canceling changes the domain. Simplifying (x² − 4)/(x − 2) to x + 2 is valid for x ≠ 2 only. The original function has a hole at x = 2 even though the simplified version does not.

A filled dot overrides the curve. If a graph shows a smooth line approaching x = 2 at height 5 but a solid dot at (2, 1), the function is discontinuous there. The limit is 5, the value is 1, and they disagree.

Oscillation defeats the limit. f(x) = sin(1/x) near x = 0 never settles on a value — it swings between −1 and 1 infinitely often. The limit does not exist, so the discontinuity is non-removable.

Working through these cases builds the pattern recognition that makes the three tests fast: check the domain first, compare one-sided limits at any suspicious point, then confirm the value matches.

Key Takeaways

• A function is continuous at x = c only if all three hold: f© exists, lim(x→c) f(x) exists, and the two are equal.
• Continuity on an interval means continuity at every point inside it, plus one-sided continuity at closed endpoints.
• Graphically, holes, jumps, and vertical asymptotes are the three break types; holes are removable, jumps and asymptotes are not.
• To check algebraically, compare one-sided limits — they must agree on a finite value — then confirm that value equals f©.
• A 0/0 result usually signals a removable hole; factor or rationalize, then substitute again.