2 Limits and Continuity

A progressive guide to interpreting, calculating, and applying limits, asymptotes, continuity, and the Intermediate Value Theorem.

Understanding the meaning of a

A describes nearby behavior rather than necessarily reporting the function's value at the target. The notation

lim⁡x→af(x)=L\lim_{x\to a}f(x)=L

means that f(x)f(x) approaches LL as xx approaches aa. The value f(a)f(a) may equal LL, differ from LL, or be undefined.

For example, consider

f(x)=x2−1x−1.f(x)=\frac{x^2-1}{x-1}.

Factoring gives

f(x)=(x−1)(x+1)x−1=x+1(x≠1).f(x)=\frac{(x-1)(x+1)}{x-1}=x+1\qquad(x\ne1).

Therefore,

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 function is undefined at x=1x=1, but its nearby values approach 22. This produces a , represented graphically by a hole at (1,2)(1,2).

Takeaway: Separate the question “What value does the function have at the point?” from the question “What value do nearby function values approach?”

Evaluating limits

Limits can be estimated with a table, interpreted from a graph, or found exactly with algebra.

A numerical table uses input values on both sides of the target. For f(x)=x2+1f(x)=x^2+1, values near x=2x=2 approach 55, suggesting

lim⁡x→2(x2+1)=5.\lim_{x\to2}(x^2+1)=5.

Numerical evidence is useful for estimation, but rounded values or inputs that are not sufficiently close may hide important behavior.

From a graph, trace the function toward the target from the left and from the right. If both sides approach the same height, the values agree and the two-sided exists. A hole does not prevent a from existing.

For exact algebraic evaluation, use these strategies:

  • Substitute directly when the function is continuous at the target. For example,
    \

    lim⁡x→3(2x2−x+4)=2(3)2−3+4=19.\lim_{x\to3}(2x^2-x+4)=2(3)^2-3+4=19.
  • If substitution produces 00\frac{0}{0}, factor and cancel common factors before substituting. For example,
    \

    lim⁡x→2x2−4x−2=lim⁡x→2(x+2)=4.\lim_{x\to2}\frac{x^2-4}{x-2}=\lim_{x\to2}(x+2)=4.
  • For radicals, multiply by the conjugate. For example,
    \

    lim⁡x→0x+1−1x=lim⁡x→01x+1+1=12.\lim_{x\to0}\frac{\sqrt{x+1}-1}{x}=\lim_{x\to0}\frac{1}{\sqrt{x+1}+1}=\frac{1}{2}.

laws permit sums, differences, constant multiples, products, quotients, and powers to be evaluated from the corresponding individual limits when those limits exist. For a quotient, the limiting denominator must be nonzero.

Takeaway: Use tables and graphs to build intuition, but use algebraic simplification when an exact value is required.

Comparing behavior from both sides

A records behavior from one direction:

lim⁡x→a−f(x)\lim_{x\to a^-}f(x)

uses values less than aa, while

lim⁡x→a+f(x)\lim_{x\to a^+}f(x)

uses values greater than aa. A two-sided exists exactly when both one-sided limits exist and are equal:

lim⁡x→af(x)=L⟺lim⁡x→a−f(x)=lim⁡x→a+f(x)=L.\lim_{x\to a}f(x)=L\quad\Longleftrightarrow\quad\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=L.

Consider

f(x)={x+1,x<2,5−x,x≥2.f(x)=\begin{cases}x+1,&x<2,\\5-x,&x\ge2.\end{cases}

The one-sided limits are

lim⁡x→2−f(x)=3,lim⁡x→2+f(x)=3.\lim_{x\to2^-}f(x)=3,\qquad\lim_{x\to2^+}f(x)=3.

Because they agree, the two-sided is 33, even though the formula changes at x=2x=2. By contrast, if the left-hand is 22 and the right-hand is 55, the two-sided does not exist.

Takeaway: Always compare the left and right sides before assigning a two-sided .

Infinite limits and vertical behavior

An infinite describes unbounded behavior; ∞\infty and −∞-\infty are not finite real-number values. For example,

lim⁡x→0+1x=∞,lim⁡x→0−1x=−∞.\lim_{x\to0^+}\frac{1}{x}=\infty,\qquad\lim_{x\to0^-}\frac{1}{x}=-\infty.

Since the one-sided behaviors differ, the two-sided at x=0x=0 does not exist. The line x=0x=0 is a .

A denominator that becomes zero may signal a , but cancellation must be checked first. The expression

x2−1x−1=x+1(x≠1)\frac{x^2-1}{x-1}=x+1\qquad(x\ne1)

has a hole at x=1x=1, not a . In contrast, f(x)=1x−1f(x)=\frac{1}{x-1} becomes unbounded near x=1x=1, so x=1x=1 is a .

Takeaway: Distinguish a hole caused by a canceled factor from unbounded behavior caused by a remaining denominator factor.

Limits at infinity and end behavior

Limits as xx tends to ∞\infty or −∞-\infty describe end behavior:

lim⁡x→∞f(x)andlim⁡x→−∞f(x).\lim_{x\to\infty}f(x)\qquad\text{and}\qquad\lim_{x\to-\infty}f(x).

For example,

lim⁡x→∞(3+4x)=3,lim⁡x→−∞(3+4x)=3.\lim_{x\to\infty}\left(3+\frac{4}{x}\right)=3,\qquad\lim_{x\to-\infty}\left(3+\frac{4}{x}\right)=3.

Thus, y=3y=3 is a .

For a rational function f(x)=p(x)q(x)f(x)=\frac{p(x)}{q(x)}, compare the degrees of the numerator and denominator:

  • If deg⁡p<deg⁡q\deg p<\deg q, the is y=0y=0.

  • If deg⁡p=deg⁡q\deg p=\deg q, the is the ratio of the leading coefficients.

  • If deg⁡p>deg⁡q\deg p>\deg q, there is generally no ; polynomial division may reveal a slant or higher-degree polynomial asymptote.

For example,

lim⁡x→∞2x2+1x2−3=2,\lim_{x\to\infty}\frac{2x^2+1}{x^2-3}=2,

because dividing numerator and denominator by x2x^2 gives

2+1x21−3x2→2+01−0=2.\frac{2+\frac{1}{x^2}}{1-\frac{3}{x^2}}\to\frac{2+0}{1-0}=2.

An asymptote describes limiting behavior, so a graph may cross it.

Takeaway: Degree comparison is a fast way to predict the end behavior of a rational function.

and discontinuities

at x=ax=a requires three conditions:

  1. f(a)f(a) is defined.

  2. lim⁡x→af(x)\lim_{x\to a}f(x) exists.

  3. lim⁡x→af(x)=f(a)\lim_{x\to a}f(x)=f(a).

Equivalently,

lim⁡x→af(x)=f(a).\lim_{x\to a}f(x)=f(a).

A occurs when the exists but the function is missing a value or has the wrong value. A jump discontinuity occurs when the one-sided limits exist but are unequal. An infinite discontinuity occurs when the function becomes unbounded near the point.

Polynomials are continuous everywhere. Rational functions are continuous wherever their denominators are nonzero, and exponential, logarithmic, trigonometric, and root functions are continuous on their domains.

For a piecewise function, test the boundary between formulas. If

f(x)={x2,x<1,2x−1,x≥1,f(x)=\begin{cases}x^2,&x<1,\\2x-1,&x\ge1,\end{cases}

then

lim⁡x→1−f(x)=1,lim⁡x→1+f(x)=1,f(1)=1.\lim_{x\to1^-}f(x)=1,\qquad\lim_{x\to1^+}f(x)=1,\qquad f(1)=1.

All three conditions hold at x=1x=1.

Takeaway: To test at a boundary, compare the left-hand , right-hand , and assigned function value.

Using to guarantee values

The connects with guaranteed existence. If a function is continuous on the closed interval [a,b][a,b], it takes every value between f(a)f(a) and f(b)f(b). In particular, if the endpoint values have opposite signs, there is at least one c∈(a,b)c\in(a,b) such that f(c)=0f(c)=0.

For

f(x)=x3−x−1,f(x)=x^3-x-1,

holds because this is a polynomial. The endpoint values are

f(1)=−1,f(2)=5.f(1)=-1,\qquad f(2)=5.

Since zero lies between −1-1 and 55, the theorem guarantees a number c∈(1,2)c\in(1,2) for which

f(c)=0.f(c)=0.

The theorem establishes that a solution exists even when an exact algebraic expression is unavailable. It does not by itself identify the solution or prove that the solution is unique.

Takeaway: prevents a function from skipping intermediate values, making sign changes a powerful existence test.