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With regard to the question:$\lim_{x\to a^{-}} f(x) =\lim_{x\to a^{+}} f(x) =L$

theorem :Suppose that $f : A \to \mathbb{R}$, where$ A ⊆ \mathbb{R}$, and $c \in \mathbb{R} $ is an accumulation point of $\{ x\in A : x > c\}$ and $ c$ is not an accumulation point $\{ x\in A : x < c\}$ Then:

$$\lim_{x\to c} f(x)=L$$

if and only if :

$$\lim_{x\to c^+} f(x)=L$$

is the theorem correct ? if correct . prove .

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user402750
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1 Answers1

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The implication $$\lim_{x\to c}f(x)=L\implies\lim_{x\to c^+}f(x)=L$$ is obvious, so let's show the converse.

To begin, select $\delta_1>0$ such that $A\cap(c-\delta_1,c)=\emptyset$. This choice is possible because $c$ is not an accumulation point of $\{x\in A:x<c\}$.

Now let $\epsilon>0$. We know that there exists some $\delta_2>0$ such that if $x\in A$ and $0<x-c<\delta_2$ then $|f(x)-L|<\epsilon$.

Then let $\delta=\min\{\delta_1,\delta_2\}$. Now, if $x\in A$ and $0<|x-c|<\delta$ then $x>c$ (because no point of $(c-\delta,c)$ is in $A$). This implies $0<x-c<\delta\le \delta_2$. Thereforee, $|f(x)-L|<\epsilon$, q.e.d.

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