Continuity and types of discontinuity

Nearby readings can disagree with the value at the point

A function gives f(x)=x+2 for every real x≠0, but separately assigns f(0)=5. Its values at −1/4 and 1/4 are 7/4 and 9/4. As inputs approach zero, the line’s outputs approach two; the value at zero is five.

The difference is exact: for every x≠0, |f(x)−2|=|x|. Any requested output tolerance ε>0 is met by all nonzero inputs with |x|<ε. Thus the limit at zero is 2, although f(0)=5. The hollow marker at (0,2) records the line’s omitted point; the filled marker at (0,5) is the assigned value.

Continuity at zero asks whether nearby outputs approach f(0), not merely whether some limit exists. Here they do not. Changing just f(0) to 2 makes the function the uninterrupted line x+2. The nearby rule and its limit stay the same; the point value changes.

Keep the domain, point and approach separate

The published preparation links practise function domains, coordinate graphs and absolute-value distances. Limit notation needed here is defined locally. First compare the line, jump and endpoint cases; return to the algebra and composition proofs after those differences are clear.

Write f:D→ℝ for a real-valued function on a set D of allowed real inputs. Each x∈D has one finite real output. Membership a∈D is the first question before discussing continuity at a. A formula may allow inputs on one side of a without allowing the other side.

An accumulation point of D has allowed inputs different from a arbitrarily close to a: for every r>0, some x∈D satisfies 0<|x−a|<r. A limit at a uses only these punctured inputs, whether or not a itself is in D. An available left approach additionally requires such inputs x<a for every r; a right approach uses x>a.

For a finite real L, lim(x→a; x∈D) f(x)=L means: for every ε>0 there is δ>0 such that every x∈D with 0<|x−a|<δ satisfies |f(x)−L|<ε. We require a to be an accumulation point when stating this limit. Two-sided limit language requires both approach sides; a domain-relative limit may have only one available side.

A tolerance must control every allowed nearby input

For a∈D, f is continuous at a relative to D if for every ε>0 there is δ>0 such that every x∈D with |x−a|<δ satisfies |f(x)−f(a)|<ε. The point x=a is included: its output error is zero. Epsilon is an arbitrary requested output tolerance, and delta is a sufficient input radius.

The radius may depend on ε and on a. One sufficient radius at one tolerance is not the complete continuity promise. Several sampled inputs cannot verify every allowed input or every positive tolerance. A proof supplies the necessary estimate.

For f(x)=2x+1 at a=0, |f(x)−f(0)|=2|x|. Choosing δ=ε/2 works for every ε>0. If ε=1, the choice δ=1/2 works because |x|<1/2 implies 2|x|<1. Equality on the strip boundary is excluded by the strict inequality.

The larger proposed radius δ=1 fails for ε=1: x=3/4 is inside that input strip, but the output error is 3/2. This refutes that particular radius, not the continuity of the function. The smaller radius still works. The diagram compares the exact input strip with the required output band.

At an accumulation point, continuity joins two tests

When a∈D is an accumulation point, f is continuous there if and only if its domain-relative finite limit exists and equals f(a). Continuity already controls every nearby input, so it controls the punctured inputs used by the limit. Conversely, a limit equal to f(a) controls all punctured inputs; adding x=a changes nothing because its error is zero.

A finite domain-relative limit is unique. If L≠M were both limits, set d=|L−M| and request error d/3 from each. Use the smaller positive radius and an allowed punctured input, which exists by accumulation. The triangle inequality would give d≤|L−f(x)|+|f(x)−M|<2d/3, a contradiction.

The absolute-value inequality used here follows from −|u|≤u≤|u| and −|v|≤v≤|v|: their sum lies between −(|u|+|v|) and |u|+|v|, so |u+v|≤|u|+|v|. It controls errors; it does not turn a finite sample list into a limit proof.

In the opening case, a=0 belongs to D and the limit is 2, but f(0)=5, so continuity fails. If zero is excluded instead, there is no value f(0) to test. Under our domain-relative convention, continuity at that missing input is not a property of the original function; we can ask whether it has a continuous extension there.

Use the inputs the function actually permits

For f(x)=√x on D=[0,4], zero has only a right approach. For any ε>0, choosing δ=ε² gives 0≤x<δ ⇒ |√x−0|<ε. Therefore f is continuous at zero relative to its domain. There is no real-domain left approach to compare, and its absence is not a jump.

An isolated point is different: some radius leaves no other domain input near it. On D={0}∪[1,2], assign f(0)=7 and f(x)=x+2 on [1,2]. With δ=1, the only permitted input satisfying |x|<δ is zero itself. Its error from f(0) is zero for every ε>0, so continuity at zero holds.

That isolated zero is not an accumulation point. Our punctured-limit definition is therefore not invoked there; continuity still has its direct meaning. The limit-equals-value test must not be used without its accumulation hypothesis.

A function is continuous on its domain when it is continuous at every point of that domain. The function 1/x is continuous on ℝ\{0}, even though its drawing has two separated branches. A point outside the domain does not create a failed continuity test at a point inside it.

Included interval endpoints use the inward approach

On an open interval (u,v), continuity means continuity at every point in that interval. For the restricted function on [u,v], with u<v, the left endpoint u needs its right-hand limit to equal f(u); the right endpoint v needs its left-hand limit to equal f(v). Interior points have both approach sides.

On [u,v), test the included endpoint u from the right; there is no assigned endpoint value at v to test in this restricted domain. On (u,v], test v from the left. Restricting a function can remove an approach side. Always state whether continuity is relative to the restricted interval or to a larger original domain.

For √x on [0,4], zero was proved continuous from the right. At any a>0 in this interval, |√x−√a|=|x−a|/(√x+√a)≤|x−a|/√a. The choice δ=ε√a controls every allowed input. At a=4 this is a left-hand endpoint estimate; at interior a both sides are allowed. Hence this restricted function is continuous on [0,4].

For 1/x at a≠0, first require |x−a|<|a|/2. Then |x|>|a|/2 and |1/x−1/a|≤2|x−a|/|a|². Thus δ=min(|a|/2,ε|a|²/2) proves continuity at every original domain point. It does not assign a value at zero or connect the two intervals across zero.

A finite approach selects the only possible repair

At an accumulation input with a finite nearby limit L, a wrong assigned value is a removable discontinuity: replacing that value by L makes the function continuous there. If the input is excluded, adding the value L gives a continuous extension. This is the sense in which a formula can have a removable hole outside its original domain.

For the family f(x)=x+2 when x≠0 and f(0)=k, the limit is 2 for every real k. Only k=2 can give continuity, by uniqueness. If k≠2, let d=|k−2|>0. For any proposed radius, choose 0<|x| smaller than both that radius and d/2. Then |f(x)−k|≥d−|x|>d/2, so the tolerance d/2 cannot be met.

For the original rational formula r(x)=(x²−1)/(x−1), D=ℝ\{1}. Factoring gives r(x)=x+1 for x≠1, so |r(x)−2|=|x−1|. Adding r(1)=2 extends it continuously. Cancellation proves an equality at allowed inputs; it does not retroactively change the original denominator or domain.

Changing one value cannot join a jump

Let s(x)=−1 for x<0 and s(x)=1 for x>0, with any finite assigned s(0)=k. The left-hand limit at zero is −1 and the right-hand limit is 1, because each branch is constant. Both approaches exist, but their finite values differ. This is a jump discontinuity.

If s were continuous at zero on ℝ, its limit would equal k, and restriction to each side would force both side limits to equal k. Their distinct values make this impossible. Altering just the point value cannot alter either punctured branch.

With k=1, the same rule restricted to D=[0,2] is constantly one. Its right limit at zero equals its point value, so it is continuous there relative to this smaller domain. This is a domain restriction, not a repair of the jump in the original full-domain function.

An infinite approach is not a finite endpoint value

For h(x)=1/x when x≠0, the right approach at zero is +∞ and the left approach is −∞. For every M>0, δ=1/M makes 0<x<δ imply h(x)>M, and −δ<x<0 imply h(x)<−M. These statements describe unbounded behaviour, not a real-number limit equal to an infinity symbol.

Assigning h(0)=0 makes the function defined on ℝ but not continuous at zero. Any radius contains a positive x<min(δ,1/2), giving h(x)>2 and an error from zero larger than one. More generally, no finite point value k repairs the break: use the positive growth bound with M=|k|+1, which forces |h(x)−k|>1.

This is an infinite discontinuity for the assigned full-domain function. For the unextended function on ℝ\{0}, zero remains outside its domain and is an infinite singularity of its formula. The graph window clips at y=±5; arrows show continuation beyond it. The branches neither stop there nor flatten at those plotted heights.

A bounded oscillation is a different failure

The common removable, jump and infinite categories do not exhaust all failures. Define f(x)=0 for x≤0 and f(x)=(−1)^⌊1/x⌋ for x>0. Here ⌊t⌋ is the greatest integer not exceeding t. On 1/(n+1)<x≤1/n, for a positive integer n, the value is (−1)ⁿ. The function is bounded between −1 and 1.

At x=1/(2n) its value is 1; at x=1/(2n+1) its value is −1. For any radius δ>0, choose a positive integer n>1/(2δ); both kinds of positive input then lie in (0,δ). If a finite right limit L existed, tolerance 1/2 would require both values to lie within 1/2 of L. But 2=|1−(−1)|≤|1−L|+|L−(−1)|<1 is impossible.

The right-hand finite limit therefore does not exist. There is no infinite right limit either, since the outputs stay bounded. The left limit is zero, but the full function is discontinuous at zero. This is neither a jump with two finite side limits nor a removable finite-limit hole.

Only the first five right-hand bands are drawn; infinitely many finer bands remain in the shaded interval near zero. That drawing is a finite illustration. The preceding two families of exact inputs prove the failure, including bands too small to see.

Combine continuous functions with their domain guards

Suppose f and g are continuous at a∈D on a common domain, and c is a fixed real number. Their sum, difference, scalar multiple and product are continuous there. For a requested error ε, ask each summand for error less than ε/2 and use the smaller radius. The triangle inequality controls the sum or difference. For cf, c=0 is constant; otherwise ask f for error ε/|c|.

For the product, first ensure |g(x)−g(a)|<1, so |g(x)|<|g(a)|+1. The identity f(x)g(x)−f(a)g(a)=g(x)[f(x)−f(a)]+f(a)[g(x)−g(a)] splits the error. Request |f(x)−f(a)|<ε/[2(|g(a)|+1)] and |g(x)−g(a)|<min(1,ε/[2(|f(a)|+1)]). The smaller radius makes each contribution less than ε/2, including zero point values.

For division, additionally require g(a)≠0. Continuity gives a neighbourhood where |g(x)−g(a)|<|g(a)|/2, hence |g(x)|>|g(a)|/2>0. For the reciprocal, |1/g(x)−1/g(a)|≤2|g(x)−g(a)|/|g(a)|². Request the smaller error min(|g(a)|/2,ε|g(a)|²/2); multiplying by continuous f then proves the quotient rule.

Constants and x are continuous directly; finite repeated products and sums therefore make every polynomial continuous. A rational formula is continuous where its original denominator is nonzero. If the denominator vanishes, this rule supplies no conclusion: a separate extension may remove a hole, or an unbounded break may persist.

For f(x)=x and g(x)=x+2 at zero, f+g=2x+2, fg=x(x+2), and f/g=x/(x+2) have point values 2,0,0 and are continuous there. The quotient’s original domain excludes −2, which is outside the displayed window. None of these conclusions assumes that successful sampled products prove a general law.

The inner output must enter the outer function’s domain

Let f:D→E be continuous at a∈D, and let g:E→ℝ be continuous at b=f(a) relative to E. Then (g∘f)(x)=g(f(x)) is continuous at a. The order matters: f acts first. The inclusion f(D)⊆E makes every composite output defined. A local version needs this inclusion only for allowed inputs sufficiently near a.

For any requested ε>0, continuity of g at b supplies a tolerance η>0 such that y∈E and |y−b|<η imply |g(y)−g(b)|<ε. Continuity of f supplies δ>0 making |f(x)−b|<η for allowed |x−a|<δ. Substitution y=f(x) proves the composite estimate. The outer test includes y=b, so an inner function that equals b exactly causes no missing case.

Take f(x)=x² on ℝ and g(u)=√u on E=[0,∞). The inner function is continuous by the product proof and always feeds a nonnegative input to g. The square-root estimates proved earlier work on this outer domain, including its endpoint zero. The composition is √(x²)=|x| and is continuous even at zero.

The outer hypothesis cannot be replaced by knowing only its punctured limit. If f(x)=0 for all x and g(u)=u+2 for u≠0 while g(0)=5, the outer punctured limit at zero is 2, but every actual composite output is g(0)=5. The composite is the continuous constant five. This also shows that the stated hypotheses are sufficient, rather than necessary in every individual example.

Predict, then check the point and every available side

Choose a supplied domain, formula and target input below. Predict continuity or whether a proposed tolerance radius works. Before reveal, the input rule, point markers and coordinate scale remain visible. Reveal the exact side-limit and domain record, nearby fractions and the reason the conclusion follows.

Hollow markers show omitted rule endpoints; filled markers show included values. Side availability follows the domain, not a blank part of the plotted window. Dashed factor curves, solid result curves and numbered coordinate ticks distinguish the algebra comparisons without relying on colour alone.

A finite table checks arithmetic, while the section’s estimate or contradiction establishes the universal conclusion. Keep the original domain when evaluating a formula, name any extension or restriction, and test only the approach sides that exist. Continuity at each point does not require one input radius to work for every target point.

Worked examples

Try it yourself

1. For f(x)=x+2 at x≠0 and f(0)=−1, state the limit at zero and test continuity there.

   Show the answer and reasoning

   The finite limit is 2, while f(0)=−1; continuity fails with a removable discontinuity.

   The nearby error from 2 is |x|, independent of the separate point assignment. Continuity at this accumulation point requires that value to equal 2.

2. The original formula r(x)=(x²−1)/(x−1) has domain ℝ\{1}. Is one a point of its domain, and what continuous extension is possible?

   Show the answer and reasoning

   One is excluded. Adding r(1)=2 gives the unique continuous extension there.

   The original function is continuous at every point of its domain. Its removable formula hole is repaired by the nearby limit 2, not by claiming the cancelled formula had originally defined one.

3. For f(x)=2x+1 at zero and ε=1, does δ=1 work? Does its failure refute continuity?

   Show the answer and reasoning

   No: x=3/4 gives error 3/2. Continuity still holds with δ=1/2.

   The failed candidate has an allowed input within its strip whose error exceeds the tolerance. Continuity asks for some sufficient radius for each tolerance, not for every proposed radius.

4. For √x restricted to [0,4], which side is available at four, and what radius controls error ε>0 there?

   Show the answer and reasoning

   Only the left approach is available; δ=2ε is sufficient.

   The target is included and the error is |x−4|/(√x+2)≤|x−4|/2 for every allowed input. The domain supplies no right-side inputs beyond four.

5. On D={0}∪[1,2], set f(0)=7 and f(x)=x+2 otherwise. Is f continuous at zero, and may the punctured-limit criterion be invoked?

   Show the answer and reasoning

   It is continuous at zero; the criterion is not invoked because zero is isolated.

   With δ=1, the only allowed input in |x|<δ is zero and the error is zero. There are no punctured inputs arbitrarily close to zero to support our limit definition.

6. A function has left limit −1 and right limit 1 at zero on a domain with both approaches. Can any finite assignment at zero make it continuous?

   Show the answer and reasoning

   No. The jump cannot be repaired by changing only the point value.

   Continuity would force both existing side limits to equal the assigned point value. One real number cannot equal both distinct limits.

7. Let h(x)=1/x at nonzero inputs and h(0)=0. Why does continuity fail although every point value is finite?

   Show the answer and reasoning

   The nearby outputs are unbounded; at every proposed radius some positive input gives error larger than one.

   Choose 0<x<min(δ,1/2). Then 1/x>2, whereas h(0)=0. Assigning one finite point value cannot stop the positive infinite approach.

8. For the bounded floor-function example, why is its missing right limit not an infinite limit or a jump with two finite side limits?

   Show the answer and reasoning

   Outputs remain in {−1,1} on the right, but both values occur arbitrarily close to zero, so the right finite limit is absent.

   Bounded outputs cannot exceed every positive growth threshold. The exact inputs 1/(2n) and 1/(2n+1) contradict any common finite right limit; a jump requires two existing finite side limits.

9. If f and g are continuous at a but g(a)=0, does the quotient continuity theorem apply? Compare x/x and 1/x near zero.

   Show the answer and reasoning

   The theorem does not apply. The first quotient has a removable hole with limit 1; the second has infinite one-sided behaviour.

   For x≠0, x/x=1, so adding value 1 gives a continuous extension. No finite value extends 1/x continuously at zero. The zero denominator alone does not choose between these outcomes.

10. For f(x)=x² on ℝ and g(u)=√u on [0,∞), is g∘f defined and continuous at zero?

    Show the answer and reasoning

    Yes. Its value is √(x²)=|x|, including at zero.

    Every inner value belongs to the outer domain. Both factors are continuous at the required points, with outer continuity relative to its one-sided domain at zero. The proved composition estimate applies.

11. For q(x)=x/(x+2), where is it continuous? Does the graph’s displayed interval [−1,1] change that domain?

    Show the answer and reasoning

    It is continuous on ℝ\{−2}; the plotted window does not change its domain.

    The original denominator is nonzero at every allowed input, so the proved quotient rule applies locally there. At −2 the numerator is nonzero and the original quotient is undefined; clipping a diagram is not a domain restriction.

12. A function is restricted to (0,2]. Which endpoint continuity test is required?

    Show the answer and reasoning

    The included endpoint two needs its left-hand limit to equal its value. No endpoint value at zero belongs to this restricted function.

    Every interior point also needs continuity. At an endpoint use only approaches available within the stated interval, while keeping the distinction from continuity on any larger original domain.

What to carry forward

• Continuity at a domain point controls the output error from that point’s actual value for every allowed nearby input.
• At an accumulation point, the domain-relative finite limit must equal the assigned value; isolated domain points satisfy the direct definition without that limit test.
• Included interval endpoints use inward approaches; restrictions, extensions and plotted windows change different things.
• A finite common limit gives a unique removable repair; unequal finite side limits form a jump, and an infinite branch cannot be repaired with a finite point value.
• Bounded oscillation can destroy a side limit; three familiar categories are not an exhaustive classification.
• Algebra requires a common domain and a nonzero denominator at the target; composition requires correct order, valid inner outputs and outer continuity at the actual inner value.
• Diagrams and exact samples help make predictions; estimates and contradictions establish the universal continuity promise.