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In a Calculus book, I had read the following proposition:

For a function $f:X\to \mathbb{R}$, $X\subseteq \mathbb{R}$ then $\lim\limits_{x\to x_0}f(x)$ exists if and only if $\lim\limits_{x\to x_0^+}f(x)$ and $\lim\limits_{x\to x_0^-}f(x)$ exist and $\lim\limits_{x\to x_0^+}f(x)=\lim\limits_{x\to x_0^-}f(x)$.

It has come to my attention however that this is wrong.

Let $f:\left[a,b\right] \to \mathbb{R}$, $f(x)=x$. Note that $a$ is an accumulation point (from the right) of the domain of $f$.

Obviously, $\lim\limits_{x\to a^-}f(x)$ does not exist since $a$ is not an accumulation point from the left of the domain of $f$. Using the definition of a limit of a real function (or using the fact that $f$ is continuous on $a$) we can derive that $\lim\limits_{x\to a}f(x)=a$ which is a contradiction to the proposition above.

My question is, in the proposition do we need to additionaly suppose that $x_0$ is an accumulation point from the right and the left of $X$?

If $x_0$ is an accumulation point of $X$ only from the right and $\lim\limits_{x\to x_0^+}f(x)$ exists then is it true that $\lim\limits_{x\to x_0}f(x)$ exists and $\lim\limits_{x\to x_0}f(x)=\lim\limits_{x\to x_0^+}f(x)$?

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1 Answer 1

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It's not "wrong"; it depends on your definition of limit.

The calculus textbook is working on a definition of limit that requires the function to be defined on a punctured neighborhood of the point $x_0$. Explicitly, the definition is:

$\lim\limits_{x\to x_0}f(x) = a$ if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that for all $x$, if $0\lt |x-x_0|\lt\delta$, then $|f(x)-a|\lt\epsilon$.

The definition implicitly requires there to be a $\delta_0\gt 0$ such that $(x_0-\delta_0,x_0+\delta_0)-\{x_0\}\subseteq\mathrm{dom}(f)$.

Under this definition, the highlighted statement is correct. You cannot compute $\lim\limits_{x\to a}f(x)$ because $f(x)$ is not defined on a punctured neighborhood of $a$.

For a slightly more general definition that requires only $x_0$ to be an accumulation point of the domain, the definition may be modified slightly as follows:

Let $x_0$ be an accumulation point of $\mathrm{dom}(f)$. Then $\lim\limits_{x\to x_0}f(x) = a$ if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that for all $x\in\mathrm{dom}(f)$, if $0\lt |x-x_0|\lt\delta$, then $|f(x)-a|\lt\epsilon$.

Under this definition, you can compute limits on endpoints without specifying sides, and the highlighted statement needs to be modified to apply only to points that are accumulation points from both sides.

It's a matter of definition.

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Under the first definition $\lim_{x\to a}f(x)$ does not exist while $f$ is continuous at $a$? –  SomeoneContinuous May 10 '12 at 17:28
    
@SomeoneContinuous: Again, depends on your definition of "continuous". Under the first definition, $f$ is continuous from the right at $a$, but not plain "continuous". Standard definition of "continuous on $[a,b]$" in calculus textbooks is: "continuous at every point $c\in (a,b)$, continuous from the right at $a$, and continuous from the left at $b$." –  Arturo Magidin May 10 '12 at 17:29
    
I see. What about my second question:If $x_0$ is an accumulation point of $X$ only from the right and $\lim\limits_{x\to x_0^+}f(x)$ exists then is it true that $\lim\limits_{x\to x_0}f(x)$ exists and $\lim\limits_{x\to x_0}f(x)=\lim\limits_{x\to x_0^+}f(x)$? –  SomeoneContinuous May 10 '12 at 17:31
    
@SomeoneContinuous: Yes; if you can only approach from one side, then the definition of limit devolves into the definition of one-sided limit. –  Arturo Magidin May 10 '12 at 17:32

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