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Limits & Continuity

Useful preparation: Functions, Domains & Transformations

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A limit describes nearby behavior, not necessarily the point value. As an input window shrinks, all allowed inputs near its center must force outputs near one common number. Numerical tables suggest answers; algebra or theorems justify them.

Two-sided limit
The number approached from both sides of an interior input. Left and right approaches must agree; the function need not be defined at the input itself.
Continuity at a point
The function is defined there, its relevant limit exists, and the limit equals the actual value. At a domain endpoint, use the approach available within the domain.
Removable discontinuity
A missing or incorrect point value with an existing finite limit; assigning that limit repairs continuity at the point.

Conditions before calculation

  • The quotient limit law requires the denominator's limit to be nonzero. A zero denominator requires further analysis, not division by zero.
  • Algebraic cancellation may be used near the target when the canceled factor is nonzero there; preserve the original function's domain.
  • The intermediate value theorem requires continuity on the entire closed interval; opposite endpoint signs alone are insufficient.

Why the method works

1. Make closeness precise

Epsilon controls output error. Delta must work for every permitted nearby input, excluding the center. This requirement is stronger than checking selected numerical values.

lim⁡x→af(x)=L  ⟺  ∀ε>0 ∃δ>0:0<∣x−a∣<δ ⟹ ∣f(x)−L∣<ε\lim_{x\to a}f(x)=L\iff \forall\varepsilon>0\ \exists\delta>0:\quad 0<|x-a|<\delta\ \Longrightarrow\ |f(x)-L|<\varepsilon

2. Simplify a punctured neighborhood

For a zero-over-zero indeterminate form, try factoring or rationalizing. Equality away from the center suffices to transfer the limit.

x−ax−a=1x+a(x≠a, a>0)\frac{\sqrt{x}-\sqrt a}{x-a}=\frac1{\sqrt x+\sqrt a}\qquad(x\ne a,\ a>0)

3. Separate existence from point assignment

Compute the one-sided limits first. Only when they agree can assigning that common value make the function continuous.

lim⁡x→a−f(x)=lim⁡x→a+f(x)=L,f(a)=L\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=L,\qquad f(a)=L

Worked examples

EXAMPLE 1

Evaluate the radical limit without decimal estimates.

  1. Direct substitution gives zero over zero, which does not determine the answer. The square root is defined on a neighborhood of 4.

    lim⁡x→4x−2x−4\lim_{x\to4}\frac{\sqrt x-2}{x-4}

  2. Multiply numerator and denominator by the conjugate. For nearby inputs other than 4, the common factor can be canceled.

    x−2x−4⋅x+2x+2=1x+2(x≠4)\frac{\sqrt x-2}{x-4}\cdot\frac{\sqrt x+2}{\sqrt x+2}=\frac1{\sqrt x+2}\quad(x\ne4)

  3. The new denominator tends to 4, so the quotient limit law now applies.

    lim⁡x→41x+2=14\lim_{x\to4}\frac1{\sqrt x+2}=\frac14

Conclusion: The limit is one quarter even though the original quotient is undefined at 4.

EXAMPLE 2

Choose the point value that makes the displayed piecewise function continuous.

  1. The formula away from 1 determines nearby behavior independently of k.

    f(x)={x2−1x−1,x≠1,k,x=1.f(x)=\begin{cases}\dfrac{x^2-1}{x-1},&x\ne1,\\k,&x=1.\end{cases}

  2. Factor on the punctured neighborhood. The same linear expression describes both sides.

    (x−1)(x+1)x−1=x+1(x≠1),lim⁡x→1f(x)=2\frac{(x-1)(x+1)}{x-1}=x+1\quad(x\ne1),\qquad \lim_{x\to1}f(x)=2

  3. Continuity requires the assigned center value to match this limit. Changing k cannot alter the limit because it changes only one input.

    f(1)=k=2f(1)=k=2

Conclusion: Exactly one assignment, k equal to 2, repairs the discontinuity.

Pitfalls & counterexamples

Tempting claim: If substitution gives zero over zero, the limit equals zero.

Correction: Zero over zero is an indeterminate form: different nearby formulas can produce different limits or no limit.

Test the claim: The radical quotient above tends to one quarter, while the quotient of the absolute value of x by x has unequal left and right limits at zero.

Check your understanding

If a function's left limit is 3 and right limit is 5, can changing its value at the point make it continuous?

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Compare your reasoning

No.

The two-sided limit fails to exist, and a point assignment cannot change either one-sided approach.

Explore an interactive illustration
A secant becomes a tangent

Parent curve y=x2\displaystyle y = x^{2}. Fix A at x=a\displaystyle x = a. Bring B closer to A to explore an instantaneous rate.

Watch the secant approach the tangent
0%
AB

Red secant: (a+h)2−a2h=2 a+h=3\displaystyle \frac{\left(a + h\right)^{2} - a^{2}}{h} = 2\, a + h = 3. Dashed green tangent: limit as h→0=2 a=2\displaystyle h \to 0 = 2\, a = 2. The limit avoids dividing by zero: simplify for h≠0\displaystyle h \ne 0, then let h approach zero.

Slope gap: 1 · The animation approaches h=0\displaystyle h = 0; it never divides by zero.

Connect this idea

Limits define derivatives and improper integrals. Continuity provides the hypotheses behind existence theorems used in root finding and optimization.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

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