The Precise Definition of a Limit

Our definition so far says "$f(x)$ gets close to $L$ as $x$ gets close to $a$." Close how close? This lesson makes those words exact. It uses everything from Continuity and Limits at Infinity and Horizontal Asymptotes, and it is the definition that mathematicians actually use to prove limit laws.

Idea clave

$\displaystyle\lim_{x \to a} f(x) = L$ means: for every $\varepsilon > 0$ there is a $\delta > 0$ such that $$0 < \lvert x - a \rvert < \delta \ \text{ implies } \ \lvert f(x) - L \rvert < \varepsilon.$$ The tolerance $\varepsilon$ comes first and can be as small as anyone likes; $\delta$ is your answer to it.

Picture a game. A challenger draws a horizontal band of height $\varepsilon$ above and below $L$. You must answer with a vertical band of half-width $\delta$ around $a$ so that the graph, over that band (ignoring $x = a$), stays inside the horizontal band. The limit is $L$ if you can win against every $\varepsilon$, however thin.

Shrink $\varepsilon$ and watch the $\delta$ that works shrink with it. For this line, $\delta = \frac{\varepsilon}{2}$ always works.

Por qué funciona

Why this order? "For every $\varepsilon$ there is a $\delta$" lets $\delta$ depend on $\varepsilon$: tighter tolerances get smaller windows. Reversing it to "there is a $\delta$ that works for every $\varepsilon$" would demand one window in which $f(x)$ is within every positive distance of $L$, so $f(x) = L$ exactly there. That is far stronger than a limit, and most limits would fail it.

Ejemplo resuelto: A proof for a line

Prove that $\lim_{x \to 3} (4x - 5) = 7$.

Let $\varepsilon > 0$ be given. We need $\lvert (4x - 5) - 7 \rvert < \varepsilon$.
Simplify the distance: $\lvert 4x - 12 \rvert = 4\lvert x - 3 \rvert$.
$4\lvert x - 3 \rvert < \varepsilon$ exactly when $\lvert x - 3 \rvert < \frac{\varepsilon}{4}$.
Choose $\delta = \frac{\varepsilon}{4}$. Then $0 < \lvert x - 3 \rvert < \delta$ gives $\lvert (4x - 5) - 7 \rvert = 4\lvert x - 3 \rvert < 4 \cdot \frac{\varepsilon}{4} = \varepsilon$.

Respuesta: Proved, with $\delta = \frac{\varepsilon}{4}$.

Ejemplo resuelto: A number for $delta$

For $\lim_{x \to 2} (3x + 1) = 7$ and $\varepsilon = 0.06$, what is the largest $\delta$ that works?

$\lvert (3x + 1) - 7 \rvert = 3\lvert x - 2 \rvert$.
We need $3\lvert x - 2 \rvert < 0.06$, so $\lvert x - 2 \rvert < 0.02$.

Respuesta: $\delta = 0.02$ (any smaller positive $\delta$ works too)

Ejemplo resuelto: A curve: bound the extra factor

Prove that $\lim_{x \to 2} x^2 = 4$.

$\lvert x^2 - 4 \rvert = \lvert x - 2 \rvert\,\lvert x + 2 \rvert$. The factor $\lvert x + 2 \rvert$ changes with $x$, so bound it.
Agree to keep $\lvert x - 2 \rvert < 1$. Then $1 < x < 3$, so $\lvert x + 2 \rvert < 5$.
Now $\lvert x^2 - 4 \rvert < 5\lvert x - 2 \rvert$, which is less than $\varepsilon$ when $\lvert x - 2 \rvert < \frac{\varepsilon}{5}$.
Choose $\delta = \min\left(1, \frac{\varepsilon}{5}\right)$: both conditions then hold.

Respuesta: Proved, with $\delta = \min\left(1, \frac{\varepsilon}{5}\right)$.

Error común

Choosing $\varepsilon$ after $\delta$, or writing "there is a $\delta$ such that for every $\varepsilon$." The quantifiers must read: for every $\varepsilon$, there is a $\delta$. Check your proof's first line: it should begin "Let $\varepsilon > 0$ be given," and $\delta$ should be defined in terms of $\varepsilon$, never the other way around.

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