Limits and Continuity

What it means for a function to approach a value, made precise enough to carry the weight the derivative and the integral put on it. The limit laws, the indeterminate forms they cannot settle, and continuity as the condition that lets a limit be read off by substitution.

Definition

The limit. $lim_{x \to a} f (x) = L$ means: for every $ε > 0$ there is a $δ > 0$ such that $0 < | x - a | < δ$ implies $| f (x) - L | < ε$ .

The condition $0 < | x - a |$ excludes $x = a$ itself. A limit describes the approach, not the arrival, which is why $f (a)$ may be anything at all, or undefined, without affecting $lim_{x \to a} f (x)$ .

One-sided limits. $lim_{x \to a^{-}}$ and $lim_{x \to a^{+}}$ restrict $x$ to one side. The two-sided limit exists exactly when both one-sided limits exist and agree.

Limit laws. Where $lim f$ and $lim g$ both exist:

	Law
Sum	$lim (f + g) = lim f + lim g$
Product	$lim (f g) = lim f \cdot lim g$
Quotient	$lim (f / g) = lim f / lim g$ , provided $lim g \neq 0$
Power	$lim f^{n} = (lim f)^{n}$

Each requires the constituent limits to exist. Applied where they do not, the laws prove nothing.

Continuity. $f$ is continuous at $a$ when three conditions hold together: $f (a)$ is defined, $lim_{x \to a} f (x)$ exists, and the two are equal. Continuity is what licenses evaluating a limit by substitution, and polynomials, rational functions away from their zeros, roots, exponentials, logarithms and the trigonometric functions are all continuous on their domains.

Discontinuities. A removable discontinuity has a limit that exists but fails to match the value, or where no value is assigned; redefining $f$ at the single point repairs it. A jump has one-sided limits that exist and differ, and no redefinition repairs it. An infinite discontinuity has a quotient growing without bound.

Indeterminate forms. A quotient tending to $0 / 0$ carries no information on its own: $\frac{x^{2} - 4}{x - 2} \to 4$ , $\frac{x}{x^{2}} \to \infty$ and $\frac{x^{2}}{x} \to 0$ are all of that form with different answers. The form signals that algebra is required, factoring, rationalising, or cancelling, before the laws can apply.

The squeeze theorem. If $g (x) \leq f (x) \leq h (x)$ near $a$ and $lim_{x \to a} g = lim_{x \to a} h = L$ , then $lim_{x \to a} f = L$ . It settles limits no algebra reaches, notably $lim_{x \to 0} x^{2} \sin (1 / x) = 0$ where the oscillating factor has no limit of its own.

The intermediate value theorem. If $f$ is continuous on $[a, b]$ and $N$ lies between $f (a)$ and $f (b)$ , then $f (c) = N$ for some $c \in [a, b]$ . Continuity is the whole hypothesis, and it is what makes bisection a valid root-finding method.

Limits at infinity. $lim_{x \to \infty} f (x) = L$ means $f$ can be brought within any $ε$ of $L$ by taking $x$ large enough. For a rational function the answer is decided by the leading terms.

Assumptions and scope

A limit says nothing about $f (a)$ . The definition excludes $x = a$ deliberately, which is why a function may have a limit at a point where it is undefined, and a value at a point where it has no limit.
The limit laws require the constituent limits to exist. Writing $lim (f g) = lim f \cdot lim g$ when one factor oscillates is not an application of the law, and the product may still have a limit.
An indeterminate form is not a value. $0 / 0$ , $\infty - \infty$ and $0 \cdot \infty$ signal that the laws do not apply, not that the limit fails to exist.
The squeeze theorem needs the bounds to converge to the SAME value. Bounds converging to different values, however tight, establish nothing.
The intermediate value theorem guarantees existence, not uniqueness or location. It says a root is present, not how many or where, and it fails outright for a function with a jump.
Continuity on a closed interval is needed for the extreme value and intermediate value theorems. On an open interval a continuous function may attain no maximum.

Worked material

Non-example

Limits that are not what they look like

A limit exists where the function does not. $f (x) = \frac{x^{2} - 4}{x - 2}$ is undefined at $x = 2$ : substituting gives $0 / 0$ . Yet the limit is 4. Cancelling the common factor leaves $x + 2$ for every $x \neq 2$ , and the numbers agree, at $x = 2.001$ the quotient is $4.001$ , at $x = 1.999$ it is $3.999$ .

Concluding "undefined at the point, so no limit" is the error. The definition excludes $x = a$ deliberately; the value there is irrelevant to where the function is heading.

A value exists where the limit does not. The step function with $sgn (x) = - 1$ for $x < 0$ and $+ 1$ for $x \geq 0$ has $f (0) = 1$ , perfectly well defined. But the left limit is $- 1$ and the right limit is $+ 1$ , so no two-sided limit exists. Having a value is no guarantee of having a limit.

The same shape underlies $\frac{| x |}{x}$ , which is $+ 1$ for $x > 0$ and $- 1$ for $x < 0$ . The disagreement that made $| x |$ non-differentiable at 0.

$0 / 0$ is not an answer. All three of these are $0 / 0$ as $x \to 0$ :

\frac{x}{x} \to 1, \frac{x^{2}}{x} \to 0, \frac{x}{x^{2}} \to \infty .

The form is identical and the answers are a finite nonzero number, zero, and no finite limit at all. Reporting "the limit is $0 / 0$ " states the obstacle, not the result.

The limit laws do not apply where a limit is missing. For $lim_{x \to 0} x^{2} \sin (1 / x)$ , the product law would give $0 \cdot lim \sin (1 / x)$ , but $\sin (1 / x)$ has no limit at 0, oscillating forever between $- 1$ and $1$ . The law's hypothesis fails, so the law says nothing.

The limit is nonetheless 0, established by squeezing between $- x^{2}$ and $x^{2}$ : at $x = 0.01$ the function is $- 0.0000506$ against a bound of $0.0001$ , and at $x = 0.001$ it is $0.00000083$ against $0.000001$ . A law that does not apply is not evidence that the limit fails to exist.

Tight bounds converging to different values prove nothing. Squeezing requires a common limit. Knowing $- 1 \leq \sin (1 / x) \leq 1$ near 0 is true and useless: the bounds tend to $- 1$ and $1$ , the band never narrows, and no conclusion follows.

The intermediate value theorem fails without continuity. The step function above has $f (- 1) = - 1$ and $f (1) = 1$ , and takes no value between them anywhere, in particular, never 0. The theorem's conclusion is not almost true here; it is simply false, and continuity was the entire hypothesis.

A removable discontinuity is repairable; a jump is not. Defining $f (2) = 4$ makes $\frac{x^{2} - 4}{x - 2}$ continuous at 2, because the limit existed and only the value was missing. No assignment at $x = 0$ repairs the step function, since no single number equals both $- 1$ and $+ 1$ .

Example

One limit of each kind

Substitution suffices: $lim_{x \to 2} (x^{2} - 4 x)$ . Polynomials are continuous everywhere, so the limit is the value: $4 - 8 = - 4$ . Approaching from either side confirms it, both one-sided computations giving $- 4.00000000$ . No work beyond evaluation is required, and recognising this case is what keeps the rest of the toolkit for where it is needed.

Algebra first: $lim_{x \to 2} \frac{x^{2} - 4}{x - 2} = 4$ . Substitution gives $0 / 0$ . Factoring the numerator as $(x - 2) (x + 2)$ and cancelling, legitimate because $x \neq 2$ throughout the approach, leaves $x + 2$ , whose limit is 4. The numbers agree: $4.001$ at $x = 2.001$ , $3.999$ at $x = 1.999$ .

One-sided disagreement: $lim_{x \to 0} \frac{| x |}{x}$ . From the right the quotient is $+ 1$ ; from the left it is $- 1$ . Both one-sided limits exist and they differ, so the two-sided limit does not exist. This is a jump discontinuity, and it is the reason $| x |$ has no derivative at the origin.

Squeeze: $lim_{x \to 0} x^{2} \sin (1 / x) = 0$ . The oscillating factor has no limit, so no law applies. But $- x^{2} \leq x^{2} \sin (1 / x) \leq x^{2}$ for every $x \neq 0$ , and both bounds tend to 0. At $x = 0.1$ the value is $- 0.00544$ inside a bound of $0.01$ ; at $x = 0.001$ it is $0.00000083$ inside $0.000001$ . The band collapses and carries the function with it.

At infinity: $lim_{x \to \infty} \frac{3 x^{2} + 2 x}{x^{2} - 5} = 3$ . Dividing numerator and denominator by $x^{2}$ gives $\frac{3 + 2 / x}{1 - 5 / x^{2}}$ , and the corrections vanish. The convergence is visible: $3.368$ at $x = 10$ , $3.0215$ at $x = 100$ , $3.00202$ at $x = 1000$ , $3.000002$ at $x = 10^{6}$ . Equal degrees, so the answer is the ratio of leading coefficients.

A limit that is infinite: $lim_{x \to 0} \frac{1}{x^{2}} = \infty$ . The quotient grows without bound, $100$ at $x = 0.1$ , $10^{6}$ at $x = 0.001$ . Writing $\infty$ records the manner of failure rather than naming a number: no real value is approached, which is why this is an infinite discontinuity and why $\int_{- 1}^{1} x^{- 2} d x$ diverges.

The first is the common case, and the only one needing no technique. Each of the others marks a distinct obstruction, a hole, a jump, an oscillation, an unbounded interval, an unbounded value, and each has its own method. Diagnosing which case is in front of you is most of the skill.

Common errors

Common misconception

The limit of $f$ at $a$ is just $f (a)$ , so a function undefined at $a$ has no limit there.

Related units

Connected

The Derivative (used by)
The Definite Integral (used by)

Learn this topic

Limits and Continuity

Used in

Calculus

Sources

Calculus: Early Transcendentals (2020)

Limits and Continuity

Assumptions and scope

Worked material

Limits that are not what they look like

One limit of each kind

Common errors

Related units

Connected

Learn this topic

Used in

Sources

Settings

Focus timer

Today

Last seven days

Where the time went

Recent sessions

Settings

Notifications

Your history