l'hospital rule suppose $lim_{x->a+} f(x) = lim_{x->a+} g(x) =0$ [question]

Question

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!

Answer 1

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 $\ln(x).$

Answer 2

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 $$ lim_{x\to 0+} \frac {-1}{x}= -\infty$$.