Function limits and continuity

Try

For f(x)=x² near a=1, an input change h gives output change (1+h)²−1=2h+h². The linear term is useful intuition, but a continuity proof must control the full error for every sufficiently small h. If |h|<δ, then |2h+h²|≤2|h|+h²<2δ+δ². This bound covers both positive and negative changes, not just a finite sample of plotted points.

Predict

Predict whether δ=1/2 guarantees output error below ε=1/2. The bound is 5/4, so that radius is too large for this guarantee. In fact h=1/4 is inside the input neighborhood and gives error 9/16, already larger than 1/2. A failed bound alone does not always prove failure of a proposed implication; here an explicit interior witness does. Distinguish these two ways of evaluating a candidate radius.

Experiment

Use the live experiment with f(x)=x². Change the base point, desired output tolerance and input radius separately. The exact supremum of the output error over the open δ-neighborhood is δ(2|a|+δ). A radius passes when this quantity is at most ε: the boundary value is not attained inside the open neighborhood, so equality still yields strict error<ε.

Live experiment: one radius, every nearby input

For f(x)=x², continuity at a asks for a radius δ such that every input with |x−a| < δ satisfies |x²−a²| < ε. At x=a the error is already zero, so the experiment checks the equivalent punctured condition 0 < |x−a| < δ. Choose an output tolerance ε, then try to fit the curve inside that band by changing δ.

Center a -2-1012Output tolerance ε 0.250.512Input radius δ 0.050.10.250.51Reset experiment

This radius fails: an interior counterexample exists.

Writing x=a+h gives |x²−a²|=|2ah+h²| ≤ 2|a||h|+h². For |h| < δ, this is strictly less than δ(2|a|+δ) = 1.25. Here ε=0.5. Thus a bound equal to ε still works: the interval endpoints are excluded.

For this quadratic, the bound is also the exact supremum over the open neighborhood. Approach its outward endpoint to see why. The largest valid radius is √(a²+ε)−|a| ≈ 0.224745; every smaller positive radius works too.

The red cross uses x≈1.362372, with |x−a|≈0.362372 < δ, but |x²−a²|≈0.856059 > ε. It is an interior point, so failure does not rely on testing an excluded endpoint.

Try a=0, ε=0.25 and δ=0.5. Explain why equality in the supremum does not violate the strict output inequality. Then move the center to 2 and repair the radius. The curve uses straight display segments and rounded arithmetic; the universal conclusion comes from the algebraic argument, not from sampled pixels. This bounded playground illustrates a proof for arbitrary ε>0; its menu alone is not that proof. Settings are temporary.

Choice	Must be fixed before
Base point and function	The continuity challenge
ε>0	Choosing δ
δ>0	Every allowed nearby input

The displayed samples illustrate the neighborhood. The algebraic bound, not the samples, establishes the guarantee.

Understand

Continuity at a relative to a domain D means that for every ε>0 there is δ>0 such that x∈D and |x−a|<δ imply |f(x)−f(a)|<ε. A function limit at a instead compares with a proposed L and excludes x=a. For uniqueness of a punctured limit, a should be an accumulation point of the domain.

Here is a direct proof for the square at arbitrary a. Require δ≤1, so |x+a|≤|x−a|+2|a|<1+2|a|. Choose δ=min(1,ε/(1+2|a|)). Then |x²−a²|=|x−a||x+a|<ε. This convenient sufficient radius need not be largest. Solving the exact quadratic inequality gives the largest radius √(a²+ε)−|a|, though numerical evaluation may use an algebraically equivalent expression to avoid cancellation.

Apply

Repair “a limit equals the function value by definition.” The function with f(x)=x for x≠0 and f(0)=7 has limit zero at zero but is not continuous there. Altering one isolated value changes continuity without changing a punctured limit.

Composition preserves continuity with a deliberate tolerance chain. First use the outer function’s continuity at f(a) to choose an intermediate tolerance η. Then use the inner function’s continuity at a to find δ ensuring its outputs lie within η. Reversing that order leaves no reason that the inner error meets the outer requirement.

Continue

Optional depth examines isolated domain points. A sufficiently small relative neighborhood may contain only a, making continuity automatic there. A punctured limit statement can become vacuous, so its customary uniqueness theorem assumes accumulation. Domains are part of the assertion, not background decoration.

Save one algebraic δ proof and one counterexample that defeats every δ using a fixed ε. Explain why successful finite tests support understanding but cannot replace universal neighborhood control. The next lesson turns a failed epsilon–delta statement into a sequence of violating inputs, connecting continuity directly with the sequence theory already established.