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Continuity & Discontinuities Calculator

Continuity and discontinuity classifier. Enter f(x) and a to compute f(a) and left/right limits, then label the point continuous, removable, jump or infinite—with explanation.

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f is continuous at a ⇔ f(a) exists and lim f = f(a)
Supports + - * / ^ and sin cos tan sqrt exp log pi e
Check continuity at this point

📖 Tutorial｜Continuity & Discontinuities

1. Definition
f is continuous at a iff f(a) is defined, the left and right limits exist and are equal to f(a). Otherwise a is a discontinuity.
2. Symbols
SymbolMeaning
f(a)Value at a
f(a−0)Left-hand limit
f(a+0)Right-hand limit
aPoint under study
3. How it works
  • We compute f(a) and f(a±h) with shrinking h;
  • Compare f(a) with the two one-sided limits;
  • Classify as removable, jump, infinite or oscillating.
4. Steps
  1. Enter f(x), e.g. (x^2-1)/(x-1);
  2. Enter the point a, e.g. 1;
  3. Click Calculate to classify;
  4. A removable discontinuity can be fixed by redefining f(a).
5. Example
Example: f(x)=(x²−1)/(x−1) at x=1.
Solution: f(1) is undefined but both one-sided limits equal 2, so it is a removable discontinuity; set f(1)=2 to make it continuous.
6. Pitfalls
Continuity needs f(a), left and right limits to agree exactly;
Removable gaps can be repaired, jump and infinite cannot;
Oscillating cases like sin(1/x) are hard to judge numerically.

❓ FAQ｜Continuity & Discontinuities

What is a removable discontinuity?
Both one-sided limits exist and agree, but f(a) is undefined or differs from the limit. Define f(a) equal to the limit and it becomes continuous.
Jump vs infinite discontinuity?
Jump: both one-sided limits finite but unequal. Infinite: at least one side tends to ±∞, a vertical asymptote.
What are the three continuity conditions at a?
① f(a) defined; ② both one-sided limits exist; ③ all three equal.
Why can’t sin(1/x) at 0 be classified?
It oscillates infinitely near 0 so f(0±h) does not settle; it is an oscillating discontinuity.
Does redefining f(a) always fix continuity?
Only for removable ones. Jump, infinite and oscillating discontinuities cannot be repaired this way.