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Formulating the theorem for l'hospital rule they write the following:

"Suppose the functions $f$ and $g$ differentiable on the interval $(a,b)$, and $g'(x) \ne 0$ there. Suppose that

(i) $\lim_{x \to a^+} f(x) = \lim_{x \to a^+} g(x) =0$

(ii)$\lim_{x \to a^+} \frac{f'(x)}{g'(x)} = L$"

Then they formulate the "second L'hospital rule"

Where I have to suppose that

(i) $\lim_{x \to a^+} g(x) = \pm \infty $

I don't understand what the distinction between

(i) $\lim_{x \to a^+} f(x) = \lim_{x \to a^+} g(x) =0$

and

$\lim_{x->a+} g(x) = \pm \infty $

are. How is it possible for a function to approach $\pm \infty$ when $a \to 0^+$?

Many thanks whomever might help me understand this!

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  • $\begingroup$ Do you know the proofs of the different cases of L'Hospital's rule? Read it somewhere, but in this case basically is what interval to take in order to apply the Mean value Theorem $\endgroup$ – DonAntonio Jan 14 '18 at 20:31
  • $\begingroup$ Indeed function $g(x)$ cannot approach both $0$ and $\infty$ as $x\to a^+$, but what is being described is (I suspect) two separate cases of l'Hopital's Rule. A rational function like $g(x) = 1/(x-3)$ approaches $+\infty$ as $x\to 3^+$ and approaches $-\infty$ as $x\to 3^-$. $\endgroup$ – hardmath Jan 14 '18 at 21:12
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Your question is about the possibility of $$ lim_{x\to a+}g(x)=±\infty$$ for a function as x approaches $a$ from the right.

Consider $$f(x)= \frac {1}{x}$$ and observe its behaviour as $x\to 0^+$.

Note that as $ x$ gets smaller and smaller, $\frac {1}{x}$ gets larger and larger without bound.

That is what we mean when we write $$ lim_{x\to 0+} \frac {1}{x}= +\infty$$

Now if you consider $\frac {-1}{x}$ , you will get $$ lim_{x\to 0+} \frac {-1}{x}= -\infty$$.

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  • $\begingroup$ I get that. But I don't understand why the second l'hospital rule have other stipulations for it to be applied. Is it that it treats different functions? $\endgroup$ – Aliasa Zarowny Pseudonymia Jan 14 '18 at 20:49
  • $\begingroup$ yes, and also they did not explain that both f and g have to diverge to \infty for the second L'Hospital's Rule. $\endgroup$ – Mohammad Riazi-Kermani Jan 14 '18 at 21:02
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If you define another function as $h(x)=1/f(x),$ and $f(x) \to \pm \infty,$ then $h(x) \to 0$ and the other case applies.

This has to be done to both numerator and denominator. That is, if both $f,g \to \pm \infty,$ then use that $f/g=(1/g)/(1/f)$ and then do the $0/0$ version, and I think the chain rule.

An example of approaching $-\infty$ as $x \to 0^+$ is $\ln(x).$

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