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I'm reading the second L'Hopital rule theorem and I'm having hard time understanding the beginning of the proof.

Let the functions $f$ and $g$ are defined, they have continuous derivatives in the open interval $(a,b)$ and $g'(x) \ne 0$ everywhere in the interval $(a,b)$.

Let $$\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=+\infty $$ and let $\lim_{x\to a}\frac{f'(a)}{g'(a)}$ exists in a generalised sense. Then $$ \lim_{x\to a}\frac{f(a)}{g(a)}=\lim_{x\to a}\frac{f'(a)}{g'(a)}. $$

The proof begin by saying that the assumptions for the theorem imply that $g'(x) < 0$. I really don't get that. How is it that the function's limit is $+\infty$, but it is actually decreasing? My guess is that one of the functions limit is viewed from the opposite direction but I don't get how is that given in the statement information.

Thanks in advance!

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  • $\begingroup$ Just FYI ... The condition $\lim_{x\to a} f(x)=\infty$ is completely unnecessary. In fact, $\lim_{x\to a}f(x)$ need not even exist. You should note that it is not used in the proof. $\endgroup$ – Mark Viola Sep 30 '17 at 14:49
  • $\begingroup$ Consider $g(x)=1/x$ on $(0,1).$ $\endgroup$ – zhw. Sep 30 '17 at 14:52
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I think it's because the limit is taken in $a$, which is the smaller limit of interval $[a,b]$. So $g(x)$ is converging towards $+\infty$ when $x\rightarrow a $, so x is decreasing towards a.

I don't know if I'm clear, but it is as you said that the limit is 'seen from the other direction' as the direction is from $x>a$ to $x=a$ and not increasing $x$. So if $g$ increases from $x$ to a direction, it decreases from $a$ to $x$ direction.

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Since $g'(x)$ is continuous and has no zeros in $(a,b)$, either it is positive everywhere or negative everywhere. If it is negative everywhere, then the function $f(x)$ is increasing, and thus cannot approach $+\infty$ at the left endpoint. (If $c \in (a,b)$, then $f(x) < f(c)$ for $a < x < c$, so $\lim_{x\to a} f(x)$ cannot be $+\infty$.) Thus, we conclude that $g'(x)$ is negative everywhere.

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