The ε–δ Definition

The rigorous meaning of

\lim_{x\to a} f(x) = L

: no matter how narrow the

\varepsilon

-band, there is a

\delta

-band around

a

where

f

stays inside it. Check

\delta = \varepsilon/2

for

f(x) = 2x

a = 1

L = 2

Examples & Exercises

Prove $\lim_{x \to 2} 3x = 6$ from the $\varepsilon$ - $\delta$ definition; give $\delta$ for each $\varepsilon > 0$ .

\delta(\varepsilon) = \frac{\varepsilon}{3},\qquad \lvert 3x - 6\rvert = 3\lvert x - 2\rvert

\varepsilon = 0.3

— for this scene's

2x

\delta = \varepsilon/2 = 0.15

; for the problem's

3x

the slope

3

goes in the denominator.

Show proof

Given $\varepsilon > 0$ , put $\delta = \frac{\varepsilon}{3}$ . If $0 < \lvert x - 2\rvert < \delta$ , then $\lvert 3x - 6\rvert = \lvert 3(x-2)\rvert = 3\lvert x - 2\rvert < 3\delta = \varepsilon$ . Hence $\lim_{x \to 2} 3x = 6$ by the definition.
The point is the factorization $\lvert 3x - 6\rvert = 3\lvert x - 2\rvert$ : the slope $3$ amplifies the input error, and $\delta = \varepsilon/3$ exactly cancels that amplification.
Numerically: $\varepsilon = 0.03$ gives $\delta = 0.01$ . For every $x$ with $\lvert x - 2\rvert < 0.01$ we have $\lvert 3x - 6\rvert = 3\lvert x - 2\rvert < 0.03 = \varepsilon$ .

Prove $\lim_{x \to 1} x^2 = 1$ with $\delta = \min\left(1, \frac{\varepsilon}{3}\right)$ .

\lvert x^2 - 1\rvert = \lvert x - 1\rvert \cdot \lvert x + 1\rvert < 3\delta \leq \varepsilon

\varepsilon = 0.6

\delta = 0.3

— the curve

x^2

admits no constant ratio

\delta/\varepsilon

; this scene's line does.

Show proof

$\lvert x^2 - 1\rvert = \lvert (x-1)(x+1)\rvert = \lvert x - 1\rvert \cdot \lvert x + 1\rvert$ . The factor $\lvert x + 1\rvert$ varies with $x$ , so first trap $x$ near $1$ .
If $\lvert x - 1\rvert < 1$ then $0 < x < 2$ , so $\lvert x + 1\rvert = x + 1 < 3$ . Hence $\lvert x^2 - 1\rvert < 3\lvert x - 1\rvert$ .
Given $\varepsilon > 0$ , put $\delta = \min\left(1, \frac{\varepsilon}{3}\right)$ . If $\lvert x - 1\rvert < \delta$ , both $\lvert x - 1\rvert < 1$ and $\lvert x - 1\rvert < \frac{\varepsilon}{3}$ hold, so $\lvert x^2 - 1\rvert < 3 \cdot \frac{\varepsilon}{3} = \varepsilon$ . Hence $\lim_{x \to 1} x^2 = 1$ .
Numerically: $\varepsilon = 0.3$ gives $\delta = \min(1, 0.1) = 0.1$ , and for $\lvert x - 1\rvert < 0.1$ we get $\lvert x^2 - 1\rvert < 3 \cdot 0.1 = 0.3 = \varepsilon$ .

Prove that if $f$ tends to both $L_1$ and $L_2$ at $a$ , then $L_1 = L_2$ (uniqueness of limits).

\lvert L_1 - L_2\rvert \leq \lvert L_1 - f(x)\rvert + \lvert f(x) - L_2\rvert < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon

\varepsilon = 0.1

— however small

\varepsilon

is, both candidates are trapped within

\varepsilon

and must coincide.

Show proof

Let $\varepsilon > 0$ . By the definition there are $\delta_1 > 0$ with $\lvert f(x) - L_1\rvert < \frac{\varepsilon}{2}$ when $0 < \lvert x - a\rvert < \delta_1$ , and $\delta_2 > 0$ with $\lvert f(x) - L_2\rvert < \frac{\varepsilon}{2}$ when $0 < \lvert x - a\rvert < \delta_2$ . Put $\delta = \min(\delta_1, \delta_2) > 0$ .
For any $x$ with $0 < \lvert x - a\rvert < \delta$ (in $\mathbb{R}$ such an $x$ always exists), the triangle inequality gives $\lvert L_1 - L_2\rvert \leq \lvert L_1 - f(x)\rvert + \lvert f(x) - L_2\rvert < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$ .
This holds for every $\varepsilon > 0$ . If $\lvert L_1 - L_2\rvert > 0$ , choosing $\varepsilon = \frac{\lvert L_1 - L_2\rvert}{2}$ gives a contradiction; hence $\lvert L_1 - L_2\rvert = 0$ , i.e. $L_1 = L_2$ .