foundation19 min studyLesson 7 in Mathematics

Limits and continuity: reasoning about approaching

Separate nearby behavior from a value at a point, and read continuity as an explicit condition.

On this page

What you’ll understand

  • Distinguish a limit from function evaluation.
  • Analyze a removable hole without dividing by zero.
  • State the three conditions required for continuity at a point.

Suppose a calculation asks what happens as an input approaches a special value. Direct substitution may fail even when nearby outputs have a clear destination. A limit describes that nearby behavior. It does not require the function to take the destination value at the special input, or even to be defined there. This distinction lets calculus describe instantaneous change without pretending that a small but nonzero interval is exactly zero.

Approaching is different from arriving

The statement that f(x)f(x) approaches LL as xx approaches aa means that outputs can be forced as close to LL as desired by restricting inputs sufficiently close to aa, while excluding x=ax=a itself. The output tolerance is chosen first; the required input tolerance may depend on it. This is a claim about every sufficiently close allowed input, not just a few entries in a table.

A two-sided limit requires agreement from both sides, whenever the domain has nearby points on each side. If the left side approaches one number and the right side another, there is no common two-sided limit. A finite plot can suggest a limit but cannot establish all arbitrarily small scales. Oscillation, steep variation, or numerical rounding can hide behavior that matters near the point.

lim⁡x→af(x)=L⟺∀ε>0 ∃δ>0 ∀x∈D: 0<∣x−a∣<δ⇒∣f(x)−L∣<ε\lim_{x\to a}f(x)=L\quad\Longleftrightarrow\quad \forall\varepsilon>0\ \exists\delta>0\ \forall x\in D:\ 0<|x-a|<\delta\Rightarrow |f(x)-L|<\varepsilon
D is the real function's domain and a is an accumulation point of D. ε is an output tolerance and δ an input tolerance; the punctured condition excludes the point itself.↔ Scroll wide equations horizontally.

A hole can be removed algebraically

The expression (x2−1)/(x−1)(x^2-1)/(x-1) is undefined at x=1x=1. Factoring its numerator gives (x−1)(x+1)(x-1)(x+1), so for every other input it equals x+1x+1. Canceling the factor does not retroactively define the original expression at one. It instead gives a simpler rule with the same nearby behavior, which is exactly the information a limit asks for.

This is a removable discontinuity: assign the missing value to the nearby limit and the resulting extension becomes continuous. Other failures cannot be repaired by changing a single point. A jump has incompatible one-sided limits, and an unbounded function near a point has no finite limiting value there. Identify the kind of behavior before applying a memorized simplification rule.

lim⁡x→1x2−1x−1=lim⁡x→1(x+1)=2\lim_{x\to1}\frac{x^2-1}{x-1}=\lim_{x\to1}(x+1)=2
The cancellation is valid only for x ≠ 1, which is sufficient for a limit. The equation asserts a limiting value, not that the original quotient is defined at x = 1.↔ Scroll wide equations horizontally.

Continuity ties nearby behavior to the actual value

Continuity at an interior point requires that the function be defined there, that its finite limit exist there, and that the two agree. Polynomials satisfy these conditions at every real input. Quotients of continuous functions do so wherever the denominator is nonzero. At an endpoint of a domain, continuity uses the appropriate one-sided approach rather than demanding inputs outside the domain.

Continuity is a property of the stated model, not a guarantee that it is physically accurate or well conditioned. A continuous function can change very quickly, and small input errors can still create significant output errors. Conversely, a digital decision rule can deliberately jump at a threshold. Ask whether a continuous approximation suits the question instead of treating continuity as an automatic mark of quality.

f continuous at a⟺lim⁡x→af(x)=f(a)f\text{ continuous at }a\quad\Longleftrightarrow\quad \lim_{x\to a}f(x)=f(a)
The equality includes the requirements that f(a) is defined and the finite limit exists. At a domain endpoint, use the corresponding one-sided limit.↔ Scroll wide equations horizontally.

Inspect a missing value without inventing division

For f(x)=(x2−1)/(x−1)f(x)=(x^2-1)/(x-1), inspect x = 0.9 and x = 1.1, then find the limit at one.

  1. Both selected inputs differ from one, so cancellation is valid. At x = 0.9, use f(x)=x+1f(x)=x+1 to get 1.9.

  2. At x = 1.1, the same simplification gives 2.1. The outputs lie on opposite sides of 2, consistent with approaching that value.

  3. For any nonzero offset h, f(1+h)=2+hf(1+h)=2+h. Its difference from 2 is exactly h, so choosing ∣h∣<ε|h|<\varepsilon forces output error below any positive tolerance ε.

The result

The limit is 2, although f(1) is undefined. The algebra establishes more than the two sample values alone.

Explore the model

Interactive lab

Interactive lab · function

Explore a continuous example

The controls select a and x in the exact polynomial y=ax2y=ax^2. For fixed a, move x toward a chosen location and compare output values. This family is continuous everywhere; it does not display a removable hole, discontinuous function, or floating-point error analysis.

Explore a continuous exampleThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA smooth example supports intuition about approach, but a finite animation does not prove a limit.

Check your understanding

A function has limit 3 at x = 2 but is assigned f(2) = 7. Is it continuous at 2?

Try it yourself

Let g(x)=(x2−4)/(x−2)g(x)=(x^2-4)/(x-2) for x ≠ 2. What value should be assigned at two to create a continuous extension?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Factor x2−4x^2-4 and keep the restriction x ≠ 2 during cancellation.

Compare with the solution

For x ≠ 2, factoring gives g(x)=x+2g(x)=x+2. The limit at two is therefore 4. Assigning g(2) = 4 matches that limit and produces the continuous function x + 2. Assigning any other value leaves a discontinuity.

Take it with you

  • A limit describes nearby allowed inputs, not necessarily the value at the point.
  • Algebraic cancellation can preserve nearby behavior without defining division by zero.
  • Continuity requires a defined value, an existing limit, and agreement between them.

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