possible theorem in limits??

**archidi** · January 8th 2009, 06:42 PM

Is the following a possible theorem in limits??

If a is not an accumulation point of the domain of ....... $f:A\subseteq R\rightarrow R$ then f has a limit over a

**Mathstud28** · January 8th 2009, 08:17 PM

Originally Posted by archidi

Is the following a possible theorem in limits??

If a is not an accumulation point of the domain of ....... $f:A\subseteq R\rightarrow R$ then f has a limit over a

Note: You may want to wait for a more senior member to validate this answer.

My interpretation of your question

"If $f:A\longmapsto \mathbb{R}$ with $A\subseteq\mathbb{R}$ , and if $a$ is not a limit point of $A$ then does $\lim_{x\to a}f(x)$ make sense?"

Answer to that question: No. For $\lim_{x\to a}f(x)$ to exist there must be a point $q\in\mathbb{R}$ such that for every $\varepsilon>0$ there exists a $\delta>0$ such that $d_{A}(x,a)<\delta\implies d_{\mathbb{R}}(f(x),q)=|f(x)-q|<\varepsilon\quad x\in A$ . But now suppose that $a$ is not a limit point of $A$ then we can pick an open neighborhood of $a$ lets call it $N_{\delta_1}(a)$ such that there does not exist an element of $A$ in that neighborhood. Now choose an value of $\varepsilon$ that requires $\delta<\delta_1$ and we run into a contradiction.

EDIT: I am sorry, I completely for some reason disregarded the $A\subseteq\mathbb{R}$ part. This can be made slightly simpler

By the defintion of limits in $\mathbb{R}\mapsto\mathbb{R}$ for $\lim_{x\to a}f(x)$ to exist there must exist a point $q\in\mathbb{R}$ such that for every $\varepsilon>0$ there exsits a $\delta>0$ such that $|x-a|<\delta\implies|f(x)-q|<\varepsilon~~x\in A$ . But now suppose that $a$ is not a limit point of $A$ then there exists a neighborhood $(x-\delta_1,x+\delta_1)$ such that there does not exist an element of $A$ within that neighborhood. Now if we choose an $\varepsilon$ that requires $\delta<\delta_1$ we run into a contradiction.

**archidi** · January 9th 2009, 04:21 PM

Originally Posted by Mathstud28

Note: You may want to wait for a more senior member to validate this answer.

My interpretation of your question

"If $f:A\longmapsto \mathbb{R}$ with $A\subseteq\mathbb{R}$ , and if $a$ is not a limit point of $A$ then does $\lim_{x\to a}f(x)$ make sense?"

Answer to that question: No. For $\lim_{x\to a}f(x)$ to exist there must be a point $q\in\mathbb{R}$ such that for every $\varepsilon>0$ there exists a $\delta>0$ such that $d_{A}(x,a)<\delta\implies d_{\mathbb{R}}(f(x),q)=|f(x)-q|<\varepsilon\quad x\in A$ . But now suppose that $a$ is not a limit point of $A$ then we can pick an open neighborhood of $a$ lets call it $N_{\delta_1}(a)$ such that there does not exist an element of $A$ in that neighborhood. Now choose an value of $\varepsilon$ that requires $\delta<\delta_1$ and we run into a contradiction.

EDIT: I am sorry, I completely for some reason disregarded the $A\subseteq\mathbb{R}$ part. This can be made slightly simpler

By the defintion of limits in $\mathbb{R}\mapsto\mathbb{R}$ for $\lim_{x\to a}f(x)$ to exist there must exist a point $q\in\mathbb{R}$ such that for every $\varepsilon>0$ there exsits a $\delta>0$ such that $|x-a|<\delta\implies|f(x)-q|<\varepsilon~~x\in A$ . But now suppose that $a$ is not a limit point of $A$ then there exists a neighborhood $(x-\delta_1,x+\delta_1)$ such that there does not exist an element of $A$ within that neighborhood. Now if we choose an $\varepsilon$ that requires $\delta<\delta_1$ we run into a contradiction.

NO NO the theorem is:

if a is not accumulation of the domain of a function: $f:A\subseteq R$ ------->R then ,THERE exists m such that

....................... $\forall\epsilon[\epsilon>0\rightarrow\exists r(r>0\wedge\forall x( x\epsilon A\wedge 0<|x-a|<r\rightarrow |f(x)-m|<\epsilon))]$ .

.......................or in words : given ε>0 there exists r>0 such that......

.....................if xεA and 0<|x-a|<r then |f(x)-m|<ε...........................

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