Argumentum epsilon-delta
Argumentum epsilon-delta est definitio limitis, in mathematica. Definitio dicit: lim x → a f ( x ) = b {\displaystyle \lim _{x\rightarrow a}f(x)=b} significat ∀ ϵ > 0 ∃ δ > 0   ut   0 < | x − a | < δ ⇒ | f ( x ) − b | < ϵ {\displaystyle \forall \epsilon >0\exists \delta >0\ {\text{ut}}\ 0<|x-a|<\delta \Rightarrow |f(x)-b|<\epsilon } Hoc est, si vis valorem f(x) esse prope quantitatem b, modo elege valorem x qui sit satis prope valorem a. Exempli gratia, quid sit lim x → 2 x 2 {\displaystyle \lim _{x\rightarrow 2}x^{2}} rogamus. Estne 4? Si | x 2 − 4 | < ϵ {\displaystyle |x^{2}-4|<\epsilon } , deinde scimus vel ( x 2 − 4 ) < ϵ {\displaystyle (x^{2}-4)<\epsilon } (si x2 > 4) vel ( 4 − x 2 ) < ϵ {\displaystyle (4-x^{2})<\epsilon } (si x2 < 4). Casus primus: si x 2 − 4 < ϵ {\displaystyle x^{2}-4<\epsilon } , deinde ( x + 2 ) ( x − 2 ) < ϵ {\displaystyle (x+2)(x-2)<\epsilon } , et x − 2 < ϵ x + 2 {\displaystyle x-2<{\frac {\epsilon }{x+2}}} . (Quod x est prope 2, licet dicere x + 2 > 0.) Et, si x2 > 4, scimus x > 2. Casus alter: si 4 − x 2 < ϵ {\displaystyle 4-x^{2}<\epsilon } , deinde ( 2 + x ) ( 2 − x ) < ϵ {\displaystyle (2+x)(2-x)<\epsilon } , tum 2 − x < ϵ 2 + x {\displaystyle 2-x<{\frac {\epsilon }{2+x}}} , hoc est − ( x − 2 ) < ϵ x + 2 {\displaystyle -(x-2)<{\frac {\epsilon }{x+2}}} , et hic est casus ubi x < 2. Habemus ergo: | x − 2 | < ϵ x + 2 {\displaystyle |x-2|<{\frac {\epsilon }{x+2}}} , quod est δ. Si vis x2 esse proximum 4, igitur, valorem x elege sicut quantitas δ est satis parva. Si vis habere x2 inter 3.99 et 4.01, ε = 0.01. Tum 2 − x < ϵ x + 2 = 0.01 x + 2 < x − 2 {\displaystyle 2-x<{\frac {\epsilon }{x+2}}={\frac {0.01}{x+2}}<x-2} . Si x = 2.1, 0.01 x + 2 = 1 / 410 {\displaystyle {\frac {0.01}{x+2}}=1/410} , et x 2 ≈ 4.0098 {\displaystyle x^{2}\approx 4.0098} , ut voluisti. Et limes est 4. Functio dicitur continua si, omnibus x in eius dominio, lim x → a = f ( a ) {\displaystyle \lim _{x\rightarrow a}=f(a)} .