About showing that $\Gamma'(x)\to -\infty$ as $x\to 0^+$

Question

In this pdf-file, on page 3, after the author in the middle of the page has shown that $\Gamma(x)\to \infty$ as $x\to 0^+$, the author says:

It then follows from the mean value theorem combined with the fact that $\Gamma'$ always increases that $\Gamma'(x)\to -\infty$ as $x\to 0$.

Could someone tell me how the mean value theorem is used here to conclude the limit of $\Gamma'(x)$ as $x\to 0^+$?

Answer 1

(i) The Mean Value Theorem can be used to prove this general principle: if a twice-differentiable function is convex on an interval $a\leq x\leq b$ then the tangent slope at $a$ is less than the slope of the secant line segment that connects $A=(a, f(a))$ to $B= (b, f(b))$. I'll leave the details of that proof as an exercise for you to pursue at your leisure.

In the link you provided, it was already shown that the Gamma function is convex, and also shown that

(ii) $\Gamma(x)>\frac{1}{ e x} $ for all sufficiently small positive values of $x$.

This latter inequality (ii) can be used to prove that certain secant slopes are arbitrarily steeply negative.

Then (i) implies that the tangent slope must be arbitrarily negative also.

Details added as edit.

In more detail, (as requested in your comments), note that (i) asserts that $f'(a) < \frac{f(b)- f(a)}{b-a}$. Holding $b$ fixed and letting $a\to 0^+$ deduce from (ii) that $ \frac{f(b)- f(a)}{b-a}\to \frac{ f(b)-+\infty}{b} =-\infty$ as $a\to 0^+$.

That is $\Gamma'(x)\to -\infty$ as $x\to 0^+$.