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The statement for Intermediate Value Theorem is thus:

Let $f$ denote a function that is continuous on the closed interval $[a, b]$ and suppose $f(a)\neq f(b)$. If $N$ is any number between $f(a)$ and $f(b)$, then there is at least one number $c$ in $(a,b)$ so that $f(c)=N$.

But who is to stop me if I put it like this?

Let $f$ denote a function that is continuous on the open interval $(a, b)$. Now, $\lim\limits_{x\to a^+} f(a)=L$ and $\lim\limits_{x\to b^-} f(b)=M$. Suppose, $L\neq M$. If $N$ is any number between $L$ and $M$, then there is at least one number $c$ in $(a,b)$ so that $f(c)=N$.

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    $\begingroup$ The only problem here is that $\lim_{x\to a^+}f(x)$ may not exist. For example, $1/x$ or $\sin(1/x)$ is continuous on the open interval $(0,1)$, but has no limit as $x\to0^+$. $\endgroup$ – mr_e_man Oct 10 at 20:44
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Nobody.

In fact, with $f$ as in your suggestion, define $g\colon[a,b]\to \Bbb R$, $$g(x)=\begin{cases}f(x)&a<x<b\\L&x=a\\M&x=b\end{cases}$$ Then the classical IVT applied to $g$ is equivalent to your variant applied to $f$. With this in mind, the standard formulation looks perhaps more appealing.

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  • $\begingroup$ But what if $L$ or $M$ doesn't exist? $\endgroup$ – mr_e_man Oct 10 at 20:12
  • $\begingroup$ @mr_e_man it seems like they are assumed to exist, but certainly they do not need to exist. The most general form of the statement would then be that f((a,b)) is an interval (possibly but not necessarily containing its endpoints) if f is continuous on (a,b). $\endgroup$ – Shalop Oct 10 at 22:41
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Nothing. That's a slightly more general statement which is also correct.

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Your version follows from the previous version. Let $F(x)=f(x)$ for $a <x<b$, $F(a)=L$ and $F(b)=M$. Then $F$ is continuous on $[a,b]$ and the first version applied to $F$ gives you the new version.

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The statement of intermediate value theorem is what you have writtem in the $1$st statement.But who told you that if the statement of intermediate value theorem is that one,then we cannot say that what you have written next is false.In fact if you take $I$ as any interval domain,then for any two points $x_1,x_2$ in $I$ consider restriction of $f$ on $[x_1,x_2]$,then apply intermediate value theorem,this will give you the generalized result. i.e. If $f:I\to R$ is a continuous fucntion on an interval domain $I$ then if $f$ takes 2 values at $x_1$ and $x_2$,then it will take all values between the 2 values for some $x$ lying between $x_1$ and $x_2$.

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