Does the intermediate value theorem apply to functions that are continuous on the open intervals $(a,b)\,$?
I know it's pretty vital for the theorem to be able to show the values of $f(a)$ and $f(b)$, but what if I have calculated the limits as $x \to a$ (from the right-hand side) and $x\to b$ (from the left-hand side)?
If these limits are of opposite signs, would this be sufficient to say that according to the intermediate value theorem, the function has at least one root?
Would it be a problem if the limits were $+$ or $-$ infinity?
Best Answer
Yes, the intermediate value theorem applies in all of those situations. The proof is a reduction argument: the IVT for the limiting values as $x \to a\!+$ and as $x \to b\!-$ can be reduced to the standard IVT. Here is how that reduction argument goes.
Assuming $\lim_{x \to a+} f(x)$ exists and is nonzero (including the possibility of an infinite limit), there exists $c \in (a,b)$ which is as close as you like to $a$ such that $f(c)$ and $\lim_{x \to a+} f(x)$ have the same sign. Similarly, assuming $\lim_{x \to b-} f(x)$ exists and is nonzero then there exists $d \in (a,b)$ which is as close to $b$ as you like such that $f(d)$ and $\lim_{x \to b-} f(x)$ have the same sign. One can make the choices of $c,d$ so that $c<d$. So if the two limits have opposite signs it follows $f(c)$ and $f(d)$ also have opposite signs, and one can therefore apply the ordinary intermediate value theorem to the interval $[c,d]$ to obtain $x \in [c,d] \subset (a,b)$ such that $f(x)=0$.