Limit Üzerine-1

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

$$\lim\limits_{x\to a^+}f(x)=L\Leftrightarrow \left(\forall (x_n)\in (A\cap (a,\infty))^{\mathbb{N}}\right)(x_n\to a\Rightarrow f(x_n)\to L)$$ olduğunu gösteriniz.

Not: $D(A):=\{x|x, A\text{'nın yığılma noktası}\}$

7, Mart, 7 Lisans Matematik kategorisinde murad.ozkoc (8,886 puan) tarafından  soruldu

Bu linkteki teoremin kanıtına benzer şekilde yapılabilir.

1 cevap

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Kanıt:

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

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

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

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

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

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

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

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

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

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

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

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

NOT:

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

$$\equiv$$

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

$$\equiv$$

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

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

29, Mart, 29 murad.ozkoc (8,886 puan) tarafından  cevaplandı

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

$$\lim\limits_{x\to a^-}f(x)=L\Leftrightarrow \left(\forall (x_n)\in (A\cap (-\infty,a))^{\mathbb{N}}\right)(x_n\to a\Rightarrow f(x_n)\to L)$$ teoremi de yukarıdaki teoremin kanıtına benzer şekilde yapılır.

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