The
idea of a limit
— "f(x) approaches L as
x approaches a" — is intuitive but
vague. What does "approaches" really mean? The same
ε-style precision
that pinned down convergence pins down limits. The definition, due to
Weierstrass:
\lim_{x \to a} f(x) = L \quad\Longleftrightarrow\quad \forall\, \varepsilon > 0\ \ \exists\, \delta > 0 \ \text{ s.t. }\ 0 < |x - a| < \delta \ \Longrightarrow\ |f(x) - L| < \varepsilon.
Picture two bands. You demand the output land within
\varepsilon of L — a horizontal band
(L - \varepsilon,\ L + \varepsilon). I answer with an input
tolerance \delta — a vertical band
(a - \delta,\ a + \delta) — promising that every
x in my band (except possibly
a itself, hence 0 < |x-a|) lands the
curve inside your band. The limit exists iff I can meet every
challenge \varepsilon, however small.
A linear limit: \lim_{x\to 2}(3x-1) = 5
We prove this from the definition. The strategy is universal: take an arbitrary
\varepsilon, work backward from the goal
|f(x) - L| < \varepsilon to discover the right
\delta, then write the proof forward.
Step 1 — measure the output's distance from L.
With f(x) = 3x - 1, L = 5,
a = 2:
|f(x) - L| = |(3x - 1) - 5| = |3x - 6| = 3\,|x - 2|.
Step 2 — relate it to the input's distance. The output gap is
exactly three times the input gap |x - 2|. To
force 3|x-2| < \varepsilon we need
|x - 2| < \tfrac\varepsilon3.
Step 3 — choose \delta. That last line names our
\delta outright:
\delta = \frac{\varepsilon}{3}.
Step 4 — verify forward. Suppose
0 < |x - 2| < \delta = \tfrac\varepsilon3. Then,
multiplying by 3,
|f(x) - 5| = 3\,|x - 2| < 3 \cdot \frac{\varepsilon}{3} = \varepsilon.
Whatever \varepsilon you name,
\delta = \varepsilon/3 answers it. So
\lim_{x\to 2}(3x - 1) = 5. \blacksquare
A nonlinear limit: \lim_{x\to 3} x^2 = 9
Now the output gap is not a constant multiple of the input gap, so we need one extra idea:
bound the troublesome factor near a, then take a
minimum.
Step 1 — factor the output gap. With
f(x) = x^2, L = 9,
a = 3:
|x^2 - 9| = |x - 3|\,|x + 3|.
We control |x - 3| directly through
\delta — but the stray factor
|x + 3| grows with x. We must cap it.
Step 2 — pre-commit to staying near 3. Insist
from the outset that \delta \le 1, so
|x - 3| < 1, i.e.
2 < x < 4. Then
|x + 3| < 4 + 3 = 7.
On this neighbourhood the stray factor never exceeds 7.
Step 3 — bound the output gap. Combining, whenever
|x - 3| < \delta \le 1,
|x^2 - 9| = |x - 3|\,|x + 3| < 7\,|x - 3|.
Step 4 — force it below \varepsilon. To make
7|x - 3| < \varepsilon we need
|x - 3| < \tfrac\varepsilon7. We have two demands on
|x-3| — "\le 1" (so the bound
7 is valid) and "< \tfrac\varepsilon7"
(so the gap is small). Satisfy both at once with the minimum:
\delta = \min\!\left\{\, 1,\ \frac{\varepsilon}{7} \,\right\}.
Step 5 — verify forward. If
0 < |x - 3| < \delta then
\delta \le 1 validates Step 2's bound, and
\delta \le \tfrac\varepsilon7 finishes it:
|x^2 - 9| < 7\,|x - 3| < 7 \cdot \frac{\varepsilon}{7} = \varepsilon.
So \lim_{x\to 3} x^2 = 9. The \min is the
signature move of every nonlinear ε–δ proof. \blacksquare
For a function f defined near a (not
necessarily at a),
\lim_{x \to a} f(x) = L
means: for every \varepsilon > 0 there exists
\delta > 0 such that
0 < |x - a| < \delta \ \Longrightarrow\ |f(x) - L| < \varepsilon.
The condition 0 < |x - a| excludes
x = a — the limit ignores the value (or absence) of
f exactly at a. When additionally
L = f(a), f is
continuous at a.
The ε–δ game, on screen
Here is f(x) = 3x - 1 near a = 2, where
L = 5. The horizontal orange band is your tolerance
(L - \varepsilon,\ L + \varepsilon); the vertical band is my reply
(a - \delta,\ a + \delta). Shrink
\varepsilon and the box flattens; then
tune \delta until the curve, where it crosses the
vertical band, stays trapped inside the horizontal band. The verdict turns green once the
whole curve-segment is boxed in — which happens exactly when
\delta \le \varepsilon/3, the value the proof found.
The definition also lets you refute a limit. Negating it: no value
L works iff there is some \varepsilon > 0
such that no \delta succeeds — every
punctured neighbourhood of a contains a point thrown outside the
\varepsilon-band.
Classic case: f(x) = \tfrac{|x|}{x} at
a = 0, equal to +1 for
x > 0 and -1 for
x < 0. Suppose some L were the
limit, and take \varepsilon = 1. Any
\delta-window around 0 contains
both a positive point (value +1) and a negative one
(value -1), which are 2 apart. They
cannot both lie within 1 of a single
L. So \varepsilon = 1 defeats every
\delta: the limit does not exist — matching the
one-sided limits
disagreeing.
The logical shape \forall \varepsilon\, \exists \delta is a
two-player game. You (the adversary) move first, naming
\varepsilon as small and mean as you like.
I respond with a \delta built from your
\varepsilon — in the linear case
\delta = \varepsilon/3, in the quadratic
\delta = \min\{1, \varepsilon/7\}. The limit holds iff I have a
winning strategy: a rule producing a working
\delta for every \varepsilon
you could throw.
The order matters absolutely. \delta is allowed to depend on
\varepsilon (it always does), but \varepsilon
may not depend on \delta — you commit first. Swapping
the quantifiers (\exists \delta\, \forall \varepsilon) would
demand one \delta good for all
\varepsilon at once, which is a different — and almost always
false — statement. Reading the quantifier order correctly is the whole art of analysis.