2.5 The Precise Definition of a Limit

Big idea

Every limit so far has rested on words like “close” and “approaches.” Those words are good guides and terrible definitions. They do not say how close, they do not say who chooses, and they cannot settle a disputed case. This section replaces them with a statement built entirely from inequalities, which is the version every theorem in the subject is actually proved from.

The replacement works by turning a vague standard into a challenge that must be met at every level. Someone names an accuracy: the output must land within a distance $\varepsilon$ of the proposed limit $L$. You answer with a window: a distance $\delta$ such that every input within $\delta$ of $a$, other than $a$ itself, produces an output meeting that accuracy. The claim $\lim_{x\to a} f(x) = L$ means you can answer every challenge, no matter how small the $\varepsilon$. A single unanswerable challenge sinks the claim.

Two features of the definition deserve attention before any algebra. First, the order of the quantifiers is fixed and not negotiable: the $\varepsilon$ comes first and the $\delta$ is allowed to depend on it. Almost always it does, and the dependence is the content of a proof. Second, the condition $0 < |x - a|$ is the formal version of the deletion you have been making all along. The point $a$ is excluded from the requirement, so a function with a hole or a stray value at $a$ is unaffected.

The payoff is not that you will compute limits this way. Substitution and the limit laws are far faster, and those laws are themselves proved from this definition once and then reused. The payoff is that ambiguous cases become decidable, that “does not exist” becomes something you can demonstrate rather than assert, and that the pictures you have been drawing acquire a meaning you can check.

Decoder

The limit of $f$ at $a$ is $L$ when, for every positive tolerance $\varepsilon$, there is a positive window width $\delta$ such that every input within $\delta$ of $a$, excluding $a$ itself, has its output within $\varepsilon$ of $L$.

In symbols: for each $\varepsilon > 0$ there exists $\delta > 0$ such that $0 < |x - a| < \delta$ implies $|f(x) - L| < \varepsilon$. Read the two absolute values as distances. The first measures how far the input sits from $a$, and the strict lower bound of $0$ deletes the point. The second measures how far the output sits from $L$.

Geometrically, $\varepsilon$ draws a horizontal band of half-height $\varepsilon$ around the line $y = L$, and your job is to produce a vertical strip of half-width $\delta$ around $x = a$ so narrow that the graph inside the strip never leaves the band. Shrinking the band forces a narrower strip; the definition asks that a workable strip exist for every band.

The classic mistake is reversing the quantifiers, which produces the claim that one $\delta$ works for all $\varepsilon$. That says something much stronger and almost always false. A related error is finding a $\delta$ that works and then apologizing for it not being the largest one. Any $\delta$ that works is a complete answer, and smaller values of $\delta$ work whenever a larger one does, which is why taking a minimum is always safe.

Definitions and results

The definition. $\lim_{x\to a} f(x) = L$ means: for every $\varepsilon > 0$ there exists $\delta > 0$ such that

$$ 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon $$

Reading the inequalities as intervals. $|x - a| < \delta$ says $a - \delta < x < a + \delta$, and $|f(x) - L| < \varepsilon$ says $L - \varepsilon < f(x) < L + \varepsilon$. Converting back and forth between the two notations is most of the work in a first proof.

Smaller deltas are free. If a given $\delta$ works for a given $\varepsilon$, so does any smaller positive number, since it describes a subset of the same inputs. This is why answers of the form $\delta = \min\{1, \varepsilon/5\}$ are legitimate.

The scratch work and the proof. The usual method is to start from the target $|f(x) - L| < \varepsilon$ and work backwards, factoring out $|x - a|$ until the required bound on it is visible. That derivation is scratch work; the proof itself states $\delta$, assumes $0 < |x-a| < \delta$, and deduces the conclusion forward.

Handling a nonlinear factor. When $|f(x) - L|$ factors as $|x - a|$ times an expression that still involves $x$, first restrict $\delta$ to at most some convenient number, use that restriction to bound the leftover factor by a constant $K$, and then take $\delta$ to be the minimum of the convenient number and $\varepsilon/K$.

One-sided versions. For the right-hand limit, replace the input condition by $a < x < a + \delta$; for the left-hand limit, by $a - \delta < x < a$. Everything else is unchanged.

Infinite limits. $\lim_{x\to a} f(x) = \infty$ means: for every $M > 0$ there exists $\delta > 0$ such that $0 < |x-a| < \delta$ implies $f(x) > M$. The tolerance $\varepsilon$ is replaced by a height $M$ to clear, and the logic is identical.

Showing a limit does not exist. Negating the definition: $L$ fails to be the limit when some $\varepsilon > 0$ admits no working $\delta$, that is, every window around $a$ contains an input whose output is at least $\varepsilon$ away from $L$. To show no limit exists at all, you must rule out every candidate $L$ at once.

Worked examples

From a tolerance to a window

Consider $\lim_{x\to 3} (4x - 5) = 7$. Suppose the tolerance is $\varepsilon = 0.01$. The output condition is

$$ |(4x-5) - 7| = |4x - 12| = 4|x-3| < 0.01 $$

which holds exactly when $|x - 3| < 0.0025$. So $\delta = 0.0025$ answers this challenge, and so would anything smaller.

The same computation answers every challenge at once. Given any $\varepsilon > 0$, take $\delta = \varepsilon/4$. Then $0 < |x-3| < \delta$ gives $|(4x-5)-7| = 4|x-3| < 4\delta = \varepsilon$. Every step is reversible here because the slope is constant, which is why lines are the easy case.

A curved graph needs a bound

Consider $\lim_{x\to 2} x^2 = 4$. The output distance factors:

$$ |x^2 - 4| = |x-2|\,|x+2| $$

The first factor is under your control, the second is not, so bound it. Agree in advance that $\delta$ will be at most $1$. Then $|x - 2| < 1$ puts $x$ between $1$ and $3$, so $x + 2$ lies between $3$ and $5$ and $|x+2| < 5$.

Now take $\delta = \min\{1, \varepsilon/5\}$. If $0 < |x-2| < \delta$, both restrictions are in force, so

$$ |x^2 - 4| = |x-2|\,|x+2| < \frac{\varepsilon}{5}\cdot 5 = \varepsilon $$

Check the shape of the answer. For a small $\varepsilon$ the minimum picks $\varepsilon/5$, and the $5$ records that the graph is about five times as steep near $x = 2$ as a line of slope $1$ would be. A steeper graph forces a narrower window, exactly as the picture suggests.

A limit that cannot exist

Let $f(x) = \dfrac{|x|}{x}$, which equals $1$ for every positive input and $-1$ for every negative one. Suppose some number $L$ were the limit at $0$, and take the challenge $\varepsilon = 1$.

Whatever $\delta > 0$ is offered, the punctured interval $0 < |x| < \delta$ contains both positive and negative inputs, so it contains an input with output $1$ and an input with output $-1$. If both outputs were within $1$ of $L$, then the distance between them would satisfy

$$ 2 = |1 - (-1)| \le |1 - L| + |L - (-1)| < 1 + 1 = 2 $$

which is impossible. So no $\delta$ works for $\varepsilon = 1$, and since $L$ was arbitrary, the limit does not exist.

Clearing a height instead of a tolerance

Consider $\lim_{x\to 0} \dfrac{1}{x^2} = \infty$. Given any $M > 0$, the requirement $\frac{1}{x^2} > M$ is equivalent to $x^2 < \frac{1}{M}$, that is $|x| < \frac{1}{\sqrt{M}}$.

So take $\delta = \frac{1}{\sqrt{M}}$. If $0 < |x| < \delta$ then $x^2 < 1/M$ and therefore $1/x^2 > M$. For $M = 10^6$ the window is $|x| < 10^{-3}$, which matches what a table of values shows.

Practice

The definition is a tool for justifying limits, not for finding them, so the drills below rehearse the limits the definition is about: the values you read off a graph and the values you compute with the laws. Knowing the number $L$ in advance is what makes an epsilon-delta argument possible at all, since the definition tests a proposed limit rather than producing one.

Start with graphical reading. Identifying $L$ from a picture is also the step where the horizontal band of half-height $\varepsilon$ has a visible meaning.

Then the algebraic values. These are the limits you would go on to verify formally, and computing them fluently keeps the proof work focused on the inequalities rather than on the arithmetic.

Quiz

Five items on reading limits from graphs and computing them algebraically, the values that an epsilon-delta argument is written to confirm.