Sequential continuity
About this app
Continuity via the sequential criterion: sequence xₙ → x₀ from the left, from the right, alternating — does the image sequence f(xₙ) settle into the ε-tube around f(x₀)? Six examples: parabola, jump, removable hole, oscillation sin(1/x), step, pole.
The sequential criterion
A function f is continuous at x₀ exactly when, for every sequence (xₙ) with xₙ → x₀, the image sequence satisfies f(xₙ) → f(x₀). To prove discontinuity, one sequence is enough: its images either approach a different value or do not converge at all. To prove continuity, the statement must hold for all sequences, so individual examples only give evidence. The criterion is equivalent to the ε-δ definition.
What the app shows
There are six functions: x² at x₀ = 1, the sign function sgn(x) at 0, sin(x)/x with f(0) = 0 as a removable gap, sin(1/x) at 0, the floor function ⌊x⌋ at 1 and 1/x at 0. Three sequences approach x₀ from the left, from the right and alternating, and their images appear on the graph. An ε-band around f(x₀) shows which image sequences end up inside, and a result panel reports “continuous” or “discontinuous at x₀”.
Things to try
Make ε small and push N up. For x², all three image sequences enter the band. For sgn(x) and ⌊x⌋, the sequence from one side stays outside. For sin(x)/x, every image sequence tends to 1 while f(0) is 0: the limit exists but does not match the function value. For sin(1/x) the images jump around, and for 1/x they diverge. Turn sequences off one at a time to see why one side alone is not enough.
Frequently asked questions
- What is the sequential criterion for continuity?
- f is continuous at x₀ if for every sequence xₙ → x₀ the image sequence f(xₙ) converges to f(x₀). If you find one sequence for which this fails, f is discontinuous at x₀. The criterion is equivalent to the epsilon-delta definition.
- What is a removable discontinuity?
- A point where the limit of f exists but differs from the function value, or where f is not defined. For sin(x)/x the limit as x → 0 is 1; defining f(0) = 1 makes the function continuous.
- Why can sin(1/x) not be made continuous at 0?
- As x → 0, sin(1/x) runs through all values between −1 and 1 faster and faster. There are sequences xₙ → 0 whose images approach different values or do not converge. So there is no limit, and no choice of f(0) makes the function continuous.
- Which examples does the app include?
- x², sgn(x), sin(x)/x with a removable gap, sin(1/x), the floor function ⌊x⌋ and 1/x. For each one you control ε, the number of terms N and which of the three sequences are active.
Subject: Analysis (one-dimensional)
More from Analysis (one-dimensional)
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- Bounded growth
- Units of an integral: rate times time
- Exponential growth and decay
- Optimisation: the largest rectangle
- Area between curves
- Area function is an antiderivative
- Rational functions and asymptotes
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