---
title: Limit Laws and the ε–δ Definition
module: Limits and Continuity
moduleNumber: 1
lessonNumber: 3
order: 103
summary: >
  The Limit Laws reduce a limit to arithmetic on simpler limits, and direct
  substitution settles polynomials and rational functions outright. The 0/0 forms
  that resist substitution yield to algebra or the Squeeze Theorem, and the ε–δ
  definition makes "arbitrarily close" precise as a pair of quantified
  inequalities.
topics: [Limits and Continuity]
sources:
  - book: Stewart
    ref: "Ch. 1 — Functions and Limits; §1.6 Calculating Limits Using the Limit Laws"
  - book: Stewart
    ref: "§1.7 The Precise Definition of a Limit"
draft: false
---

The [intuitive definition](/calculus/limits-and-continuity/the-limit-of-a-function)
explains what a limit means but not how to compute one without a table. The **Limit
Laws** close that gap: limits respect the arithmetic operations, so a complicated limit
assembles from trivial ones. Where the laws stall, on the indeterminate $0/0$ form,
algebra or the Squeeze Theorem takes over, and the $\varepsilon$-$\delta$ definition
replaces "arbitrarily close" with quantified inequalities that support proof.

## The Limit Laws

> **Theorem (Limit Laws).** Suppose $c$ is a constant and $\lim_{x \to a} f(x)$ and
> $\lim_{x \to a} g(x)$ both exist. Then
> $$
> \begin{aligned}
> &1.\ \lim_{x \to a}\,[f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)
> && \text{(Sum)} \\
> &2.\ \lim_{x \to a}\,[f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)
> && \text{(Difference)} \\
> &3.\ \lim_{x \to a}\,[c\,f(x)] = c \lim_{x \to a} f(x)
> && \text{(Constant Multiple)} \\
> &4.\ \lim_{x \to a}\,[f(x)\,g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)
> && \text{(Product)} \\
> &5.\ \lim_{x \to a}\,\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)},
> \ \ \lim_{x \to a} g(x) \ne 0 && \text{(Quotient)}
> \end{aligned}
> $$

Stated in words: the limit of a sum is the sum of the limits, the limit of a product
is the product of the limits, and so on, with the one caveat that a quotient needs a
nonzero denominator limit.[^stewart-laws] Repeated application of the Product Law to
$f = g$ gives a **Power Law**, and inverting a power gives a **Root Law**:

$$
\lim_{x \to a}\,[f(x)]^n = \Bigl[\lim_{x \to a} f(x)\Bigr]^n,
\qquad
\lim_{x \to a}\,\sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)},
$$

the root form requiring $\lim f(x) > 0$ when $n$ is even. Two base cases seed
everything, and both are immediate from the definition:

$$
\lim_{x \to a} c = c, \qquad \lim_{x \to a} x = a.
$$

From the base cases and the laws, a limit like $\lim_{x\to 2}(x^3 + 2x^2 - 1)$
evaluates by structure alone: powers of the limit $2$, scaled and summed, giving
$8 + 8 - 1 = 15$. Written out, each step names the law that licenses it.

> **Worked example (a limit-law chain).** Evaluate
> $\displaystyle \lim_{x \to 5}(2x^2 - 3x + 4)$, justifying each step.
>
> $$
> \begin{aligned}
> \lim_{x \to 5}(2x^2 - 3x + 4)
> &= \lim_{x \to 5} 2x^2 - \lim_{x \to 5} 3x + \lim_{x \to 5} 4 && \text{(Sum, Difference)} \\
> &= 2\lim_{x \to 5} x^2 - 3\lim_{x \to 5} x + \lim_{x \to 5} 4 && \text{(Constant Multiple)} \\
> &= 2(5^2) - 3(5) + 4 && \text{(Power, base cases)} \\
> &= 50 - 15 + 4 = 39.
> \end{aligned}
> $$

For a rational function the Quotient Law applies once the denominator's limit is seen
to be nonzero.

> **Worked example (a rational limit-law chain).** Evaluate
> $\displaystyle \lim_{x \to -2}\frac{x^3 + 2x^2 - 1}{5 - 3x}$.
>
> The denominator's limit is $\lim_{x \to -2}(5 - 3x) = 5 - 3(-2) = 11 \ne 0$, so the
> Quotient Law is valid:
> $$
> \lim_{x \to -2}\frac{x^3 + 2x^2 - 1}{5 - 3x}
> = \frac{\lim_{x \to -2}(x^3 + 2x^2 - 1)}{\lim_{x \to -2}(5 - 3x)}
> = \frac{(-2)^3 + 2(-2)^2 - 1}{11}
> = \frac{-8 + 8 - 1}{11} = -\frac{1}{11}.
> $$

### Direct substitution

That last computation is an instance of a general shortcut.

> **Theorem (Direct Substitution Property).** If $f$ is a polynomial or a rational
> function and $a$ is in the domain of $f$, then
> $$
> \lim_{x \to a} f(x) = f(a).
> $$

For these functions, evaluating the limit is nothing more than plugging in $a$.
This property characterizes the [continuous](/calculus/limits-and-continuity/continuity)
functions. Substitution disposes of any limit where the direct value is defined; the
remaining cases are those where it is not.

## Indeterminate forms

When substitution yields $0/0$, the expression is an **indeterminate form**: the
quotient's limit could be any number, and the form alone decides nothing. The value
$0$ in the denominator flags a common factor of $x - a$ shared with the numerator,
and removing it usually reveals the limit.

**Factoring.** For

$$
\lim_{x \to 1} \frac{x^2 - 1}{x - 1},
$$

substitution gives $0/0$. Factor and cancel the offending $x - 1$, legitimate because
the limit ignores $x = 1$ itself:

$$
\frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \quad (x \ne 1),
\qquad \lim_{x \to 1} (x + 1) = 2.
$$

**Rationalizing.** When a square root blocks factoring, multiply by the conjugate.
For

$$
\lim_{t \to 0} \frac{\sqrt{t^2 + 9} - 3}{t^2},
$$

multiply top and bottom by $\sqrt{t^2 + 9} + 3$:

$$
\frac{\sqrt{t^2 + 9} - 3}{t^2} \cdot \frac{\sqrt{t^2 + 9} + 3}{\sqrt{t^2 + 9} + 3}
= \frac{(t^2 + 9) - 9}{t^2\bigl(\sqrt{t^2 + 9} + 3\bigr)}
= \frac{1}{\sqrt{t^2 + 9} + 3}.
$$

Now substitution works, giving $\dfrac{1}{\sqrt{9} + 3} = \dfrac{1}{6}$.

**Combining fractions.** A quotient of fractions clears once the numerator is put over a
common denominator.

> **Worked example (a compound fraction).** Evaluate
> $\displaystyle \lim_{h \to 0} \frac{\dfrac{1}{3 + h} - \dfrac{1}{3}}{h}$.
>
> Substitution gives $0/0$. Combine the top over the common denominator $3(3 + h)$:
> $$
> \frac{1}{3 + h} - \frac{1}{3} = \frac{3 - (3 + h)}{3(3 + h)} = \frac{-h}{3(3 + h)}.
> $$
> Dividing by $h$ cancels it, legitimate for $h \ne 0$:
> $$
> \frac{1}{h}\cdot\frac{-h}{3(3 + h)} = \frac{-1}{3(3 + h)}
> \ \xrightarrow{\ h \to 0\ }\ \frac{-1}{3 \cdot 3} = -\frac{1}{9}.
> $$

The three tactics — factor-and-cancel, rationalize, and combine fractions — handle most
algebraic $0/0$ forms.

| Form after substitution | Meaning | First tactic |
| --- | --- | --- |
| a defined number | limit equals it | direct substitution |
| $0/0$ | shared factor $x - a$ | factor and cancel, or rationalize |
| (nonzero)$/0$ | vertical asymptote | one-sided limits, $\pm\infty$ |

## The Squeeze Theorem

Some limits resist algebra entirely, and for those a comparison argument works.
First, limits preserve the ordering of functions.

> **Theorem (Order).** If $f(x) \le g(x)$ for all $x$ near $a$ (except possibly at
> $a$) and both limits exist, then $\displaystyle \lim_{x \to a} f(x) \le \lim_{x \to a} g(x)$.

Trapping a function between two others that share a limit pins its limit as well.

> **Theorem (Squeeze Theorem).** If $f(x) \le g(x) \le h(x)$ for all $x$ near $a$
> (except possibly at $a$) and
> $$
> \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L,
> $$
> then $\displaystyle \lim_{x \to a} g(x) = L$.

$$
% caption: The middle curve is trapped between two bounding curves that meet at
% the same height $L$ over $a$, forcing its limit to $L$.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (-2.4,0) -- (2.6,0) node[right] {$x$};
\draw[black, ->] (0,-0.3) -- (0,3.4) node[above] {$y$};
% upper bound h
\draw[black, very thick] plot[domain=-2.1:2.1, samples=60] (\x, {1.7 + 0.42*\x*\x});
\node[black, anchor=south, font=\scriptsize] at (-1.9,{1.7+0.42*3.61}) {$h$};
% lower bound f
\draw[black, very thick] plot[domain=-2.1:2.1, samples=60] (\x, {1.7 - 0.36*\x*\x});
\node[black, anchor=north, font=\scriptsize] at (-1.9,{1.7-0.36*3.61}) {$f$};
% squeezed g wiggling between
\draw[acc, very thick] plot[domain=-2.05:2.05, samples=120]
  (\x, {1.7 + 0.10*\x + 0.20*\x*\x*sin(\x*260)});
\node[acc, anchor=west, font=\scriptsize] at (1.7,2.55) {$g$};
% target point
\fill[black] (0,1.7) circle (1.8pt);
\draw[black, dashed] (0,1.7) -- (-0.9,1.7);
\node[anchor=east, font=\scriptsize] at (-0.9,1.7) {$L$};
\node[anchor=north, font=\scriptsize] at (0,-0.05) {$a$};
\end{tikzpicture}
$$

The standard application handles an oscillation that never settles.

> **Worked example (squeezing an oscillation).** Show that
> $\displaystyle \lim_{x \to 0} x^2 \sin\frac{1}{x} = 0$.
>
> No product law applies, because $\sin(1/x)$ swings between $-1$ and $1$ infinitely
> often as $x \to 0$ and has no limit there. But $-1 \le \sin(1/x) \le 1$ for every
> $x \ne 0$, and multiplying through by $x^2 \ge 0$ preserves the inequalities:
> $$
> -x^2 \le x^2 \sin\frac{1}{x} \le x^2.
> $$
> Both outer functions tend to $0$: $\lim_{x \to 0}(-x^2) = \lim_{x \to 0} x^2 = 0$. By
> the Squeeze Theorem the middle function is trapped to the same limit, so
> $\lim_{x \to 0} x^2 \sin(1/x) = 0$.

The oscillation is real, but its amplitude is crushed to zero by the $x^2$ envelope.

$$
% caption: The envelopes $y = x^2$ and $y = -x^2$ (dashed) pinch $y = x^2\sin(1/x)$ to
% zero at the origin; the oscillation persists but its amplitude vanishes.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (-1.9,0) -- (2.0,0) node[right]{$x$};
\draw[black, ->] (0,-1.95) -- (0,2.05) node[above]{$y$};
\draw[black, very thick, dashed] plot[domain=-1.35:1.35, samples=50] (\x, {\x*\x});
\draw[black, very thick, dashed] plot[domain=-1.35:1.35, samples=50] (\x, {-\x*\x});
\draw[acc, very thick] plot[domain=0.06:1.35, samples=420] (\x, {\x*\x*sin(57.2958/\x)});
\draw[acc, very thick] plot[domain=-1.35:-0.06, samples=420] (\x, {\x*\x*sin(57.2958/\x)});
\node[black, anchor=west, font=\scriptsize] at (1.4,1.72) {$y = x^2$};
\end{tikzpicture}
$$

## The precise definition

"Arbitrarily close" and "sufficiently close" carry the intuitive definition but are
too vague to prove theorems with. The $\varepsilon$-$\delta$ definition replaces them
with quantified inequalities.

> **Definition (Precise definition of a limit).** Let $f$ be defined on some open
> interval containing $a$, except possibly at $a$ itself. Then
> $\displaystyle \lim_{x \to a} f(x) = L$ means: for every number $\varepsilon > 0$
> there is a number $\delta > 0$ such that
> $$
> \text{if } \ 0 < |x - a| < \delta \ \text{ then } \ |f(x) - L| < \varepsilon.
> $$

Here $\varepsilon$ measures a target tolerance on the output and $\delta$ a
tolerance on the input. The statement reads geometrically: for any horizontal band of
half-width $\varepsilon$ around $y = L$, there is a vertical strip of half-width
$\delta$ around $x = a$ whose graph (excluding $x = a$) lands entirely inside the
band. The clause $0 < |x - a|$ excludes $x = a$, matching the limit's indifference to
$f(a)$.

$$
% caption: For a chosen output band of half-width $\varepsilon$ around $L$, a small
% enough input strip of half-width $\delta$ around $a$ maps into the band; the band
% edges are $L \pm \varepsilon$ and the strip edges $a \pm \delta$.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{band}{HTML}{4A6FA5}
\draw[black, ->] (-0.3,0) -- (4.9,0) node[right] {$x$};
\draw[black, ->] (0,-0.3) -- (0,4.0) node[above] {$y$};
% output band (half-width epsilon; symbol kept in caption)
\fill[band!10] (0,2.05) rectangle (4.4,2.95);
\draw[black, dashed] (0,2.5) -- (4.4,2.5);
\draw[acc] (0,2.05) -- (4.4,2.05);
\draw[acc] (0,2.95) -- (4.4,2.95);
\node[anchor=east, font=\scriptsize] at (0,2.5) {$L$};
\node[band, anchor=west, font=\scriptsize] at (4.45,2.5) {output band};
% input strip (half-width delta)
\fill[acc!12] (2.0,0) rectangle (3.0,4.0);
\draw[black, dashed] (2.5,0) -- (2.5,4.0);
\node[anchor=north, font=\scriptsize] at (2.5,-0.05) {$a$};
\node[black, anchor=south, font=\scriptsize] at (2.5,4.0) {input strip};
% curve through the box intersection
\draw[very thick] plot[domain=0.3:4.2, samples=60] (\x, {2.5 + 0.9*(\x-2.5) - 0.06*(\x-2.5)*(\x-2.5)*(\x-2.5)});
\end{tikzpicture}
$$

The definition frames a contest between two parties. One names a tolerance
$\varepsilon$, however small; the other must supply a $\delta$ that keeps the output
within $\varepsilon$. The limit equals $L$ precisely when the second party can always
answer, no matter how small the first makes $\varepsilon$.

### A worked ε–δ proof

To prove $\displaystyle \lim_{x \to 3} (4x - 5) = 7$, work backward from the target to
find the $\delta$, then present the argument forward.

**Finding $\delta$.** The output tolerance is $|(4x - 5) - 7| < \varepsilon$.
Simplify the left side:

$$
|(4x - 5) - 7| = |4x - 12| = 4\,|x - 3|.
$$

This is below $\varepsilon$ exactly when $|x - 3| < \varepsilon/4$, so
$\delta = \varepsilon/4$ is the choice.

> **Proof.** Given $\varepsilon > 0$, let $\delta = \varepsilon/4$. If
> $0 < |x - 3| < \delta$, then
>
> $$
> |(4x - 5) - 7| = 4\,|x - 3| < 4\delta = 4 \cdot \frac{\varepsilon}{4} = \varepsilon.
> $$
>
> Since an answering $\delta$ exists for every $\varepsilon > 0$, the limit is
> $7$. $\blacksquare$

The structure is the same for every such proof: solve the output inequality for
$|x - a|$ to discover $\delta$, then verify.

A nonlinear function makes the last step harder. The factor multiplying $|x - a|$ is no
longer constant, so it must first be bounded by restricting $x$ to a small window, which
forces $\delta$ to be the smaller of two constraints.

> **Worked example ($\varepsilon$-$\delta$ for a quadratic).** Prove that
> $\displaystyle \lim_{x \to 3} x^2 = 9$.
>
> **Finding $\delta$.** The output tolerance is $|x^2 - 9| < \varepsilon$. Factor to
> expose $|x - 3|$:
> $$
> |x^2 - 9| = |x + 3|\,|x - 3|.
> $$
> The factor $|x + 3|$ is not constant, so bound it by first insisting $|x - 3| < 1$.
> Then $2 < x < 4$, hence $5 < x + 3 < 7$, so $|x + 3| < 7$ and
> $$
> |x^2 - 9| = |x + 3|\,|x - 3| < 7\,|x - 3|.
> $$
> This is below $\varepsilon$ once $|x - 3| < \varepsilon/7$. Both constraints hold when
> $\delta = \min\{1,\ \varepsilon/7\}$.
>
> **The proof.** Given $\varepsilon > 0$, let $\delta = \min\{1,\ \varepsilon/7\}$. If
> $0 < |x - 3| < \delta$, then $|x - 3| < 1$ gives $|x + 3| < 7$, and $|x - 3| < \varepsilon/7$
> gives
> $$
> |x^2 - 9| = |x + 3|\,|x - 3| < 7 \cdot \frac{\varepsilon}{7} = \varepsilon.
> $$
> An answering $\delta$ exists for every $\varepsilon > 0$, so the limit is $9$.

$$
% caption: Restricting $x$ to the unit window $(2, 4)$ about $3$ keeps $|x+3| < 7$, so
% $\delta = \min\{1,\ \varepsilon/7\}$ controls $|x^2-9|$.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (-0.4,0) -- (6.5,0) node[right]{$x$};
\fill[acc!12] (1,0) rectangle (5,0.95);
\draw[acc] (1,0) -- (1,0.95);
\draw[acc] (5,0) -- (5,0.95);
\foreach \px/\lab in {1/2, 3/3, 5/4} {
  \draw[black] (\px,0.12) -- (\px,-0.12);
  \node[anchor=north, font=\scriptsize] at (\px,-0.16){$\lab$};
}
\fill (3,0) circle (1.6pt);
\draw[thick, ->] (3,0.55) -- (4.7,0.55);
\draw[thick, ->] (3,0.55) -- (1.3,0.55);
\node[anchor=south, font=\scriptsize] at (3,0.64){$x$ kept near $3$};
\end{tikzpicture}
$$

A one-sided limit uses the same template with the input interval halved to one side. The
right-hand definition asks that $|f(x) - L| < \varepsilon$ whenever $a < x < a + \delta$.

> **Worked example (a one-sided $\varepsilon$-$\delta$ proof).** Prove that
> $\displaystyle \lim_{x \to 0^+} \sqrt{x} = 0$.
>
> Here $a = 0$ and $L = 0$, with $x > 0$. The output tolerance $|\sqrt{x} - 0| < \varepsilon$
> is $\sqrt{x} < \varepsilon$; squaring the positive quantities, $x < \varepsilon^2$. So take
> $\delta = \varepsilon^2$. Given $\varepsilon > 0$, if $0 < x < \delta = \varepsilon^2$, then
> $\sqrt{x} < \sqrt{\varepsilon^2} = \varepsilon$. The right-hand limit is $0$.

```algorithm
caption: Proving $\lim_{x \to a} f(x) = L$ by the $\varepsilon$-$\delta$ definition
let $\varepsilon > 0$ be given
simplify $|f(x) - L|$ to expose the factor $|x - a|$
solve $|f(x) - L| < \varepsilon$ for $|x - a|$ to obtain a bound $\delta(\varepsilon)$
choose $\delta \gets \delta(\varepsilon)$
assume $0 < |x - a| < \delta$
derive $|f(x) - L| < \varepsilon$ from the assumption
conclude the limit equals $L$
```

The Limit Laws themselves rest on this definition: the Sum Law, for instance, follows by
choosing $\delta$ so that $|f(x) - L|$ and $|g(x) - M|$ are each below $\varepsilon/2$,
whence $|(f + g)(x) - (L + M)| < \varepsilon$ by the triangle inequality. Once proved,
the laws evaluate complicated limits without returning to $\varepsilon$ and $\delta$.

[^stewart-laws]: Stewart, §1.6 — the Limit Laws (sum, difference, constant multiple, product, quotient) with the power and root laws, the Direct Substitution Property, and the Squeeze Theorem.