Gerçel tanım kümeli ve gerçel değerli fonksiyonlarda limit-$1$

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Tanım: $A\subseteq\mathbb{R}, \ f\in\mathbb{R}^A, \ a\in D(A)$  ve  $L\in\mathbb{R}$ olmak üzere

$$\lim\limits_{x\to a}f(x)=L:\Leftrightarrow (\forall \epsilon >0)(\exists \delta >0)(\forall x\in A)(0<|x-a|<\delta\Rightarrow |f(x)-L|<\epsilon)$$

Teorem: $A\subseteq\mathbb{R}, \ f\in\mathbb{R}^A, \ a\in D(A)$  ve  $L\in\mathbb{R}$ olmak üzere

$$\lim\limits_{x\to a}f(x)=L\Leftrightarrow \left(\forall (x_n)\in (A\setminus \{a\})^{\mathbb{N}}\right)(x_n\to a\Rightarrow f(x_n)\to L)$$

28, Şubat, 28 Lisans Matematik kategorisinde murad.ozkoc (8,870 puan) tarafından  soruldu
1, Mart, 1 DoganDonmez tarafından düzenlendi

İki yerde $f(x)$ unutulmuş onları ekledim.

Teşekkür ederim :-)

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Teorem: $A\subseteq\mathbb{R},$  $f\in \mathbb{R}^A, \ a\in D(A)$  ve  $L\in\mathbb{R}$ olmak üzere

$$\lim\limits_{x\to a}f(x)=L\Leftrightarrow \left(\forall (x_n)\in (A\setminus\{a\})^{\mathbb{N}}\right)(x_n\to a\Rightarrow f(x_n)\to L)$$

$-----------------------------------$

Kanıt:

$(\Rightarrow):$ $\lim\limits_{x\to a}f(x)=L,$ $(x_n)\in (A\setminus\{a\})^{\mathbb{N}}, \ x_n\to a$ ve $\epsilon>0$ olsun.

$\left.\begin{array}{r} \epsilon>0 \\ \\ \lim\limits_{x\to a}f(x)=L \end{array} \right\}\Rightarrow \begin{array}{c} \\ \\ \left. \begin{array}{r} (\exists \delta>0)(A\cap [(a-\delta,a)\cup (a,a+\delta)]\subseteq f^{-1}[(L-\epsilon,L+\epsilon)]) \\ \\ (x_n\to a)\left((x_n)\in (A\setminus\{a\})^{\mathbb{N}}\right) \end{array} \right\} \Rightarrow \end{array}$

$\begin{array}{r} \Rightarrow (\exists K\in\mathbb{N})(n\geq K\Rightarrow  x_n\in A\cap [(a-\delta,a)\cup (a,a+\delta)]\subseteq f^{-1}[(L-\epsilon,L+\epsilon)]) \end{array}$

$\begin{array}{r} \Rightarrow (\exists K\in\mathbb{N})(n\geq K\Rightarrow  f(x_n)\in f\left[A\cap [(a-\delta,a)\cup (a,a+\delta)]\right]\subseteq (L-\epsilon,L+\epsilon)) \end{array}$

$\Rightarrow (\exists K\in\mathbb{N})(n\geq K\Rightarrow f(x_n)\in (L-\epsilon,L+\epsilon)).$

$-----------------------------------$

$(\Leftarrow):$ $\lim\limits_{x\to a}f(x)\neq L$ olsun.

$\lim\limits_{x\to a}f(x)\neq L\Rightarrow (\exists \epsilon>0)(\forall\delta >0)(A\cap [(a-\delta,a)\cup (a,a+\delta)]\nsubseteq f^{-1}[(L-\epsilon,L+\epsilon)])$

$\Rightarrow (\exists \epsilon>0)(\forall n\in\mathbb{N})\left(A\cap \left[\left(a-\frac1n,a\right)\cup \left(a,a+\frac1n\right)\right]\nsubseteq f^{-1}[(L-\epsilon,L+\epsilon)]\right)$

$\Rightarrow (\exists \epsilon>0)(\forall n\in\mathbb{N})\left(\exists x_n\in A\cap \left[\left(a-\frac1n,a)\cup (a,a+\frac1n \right)\right]\right)(f(x_n)\notin (L-\epsilon,L+\epsilon))$

$\Rightarrow \left(\exists (x_n)\in (A\setminus\{a\})^{\mathbb{N}}\right)(x_n\to a)(f(x_n)\nrightarrow L).$

$-----------------------------------$

NOT:

$$\left[\left(\forall (x_n)\in (A\setminus\{a\})^{\mathbb{N}}\right)(x_n\to a\Rightarrow f(x_n)\to L)\right] \Rightarrow \left[\lim\limits_{x\to a}f(x)=L\right]$$

$$\equiv$$

$$\left[\lim\limits_{x\to a}f(x)= L\right]'\Rightarrow \left[\left(\forall (x_n)\in (A\setminus\{a\})^{\mathbb{N}}\right)(x_n\to a\Rightarrow f(x_n)\to L)\right]'$$

$$\equiv$$

$$\lim\limits_{x\to a}f(x)\neq L\Rightarrow \left(\exists (x_n)\in (A\setminus\{a\})^{\mathbb{N}}\right)(x_n\to a)(f(x_n)\nrightarrow L)$$

$-----------------------------------$

1, Mart, 1 murad.ozkoc (8,870 puan) tarafından  cevaplandı
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