L'Hospital when limit tends to infinity L'Hospital's rule states that for functions f and g which are differentiable on an open interval I except possibly at a point c contained in I, if $\lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0{\text{ or }}\pm \infty ,$ and $g'(x)\neq0$ for all x in I with x ≠ c, and  $\lim _{x\to c}{\frac {f'(x)}{g'(x)}} $ exists, then
$\lim _{x\to c}{\frac {f'(x)}{g'(x)}}=\lim _{x\to c}{\frac {f(x)}{g(x)}} $
By the definition, we know that there is nothing we can say about $\lim _{x\to c}{\frac {f(x)}{g(x)}} $
when $\lim _{x\to c}{\frac {f'(x)}{g'(x)}} $ does not exist. But i have this hypothesis that when $\lim _{x\to c}{\frac {f'(x)}{g'(x)}}=\infty $ then $\lim _{x\to c}{\frac {f(x)}{g(x)}}=\infty $ as well-- is my hypothesis correct? Or can anyone provide a counterexample?
I know this question seems like a duplicate to Question concerning L'Hospital's rule , but it doesn't give any counterexamples as stated in my above hypothesis.
Thanks!
 A: We can show under the given hypotheses that $\displaystyle \lim_{x \to c} \frac{f'(x)}{g'(x)}= \pm\infty$ implies $\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)}= \pm\infty$.
I prove this here  for $f'(x)/g'(x) \to +\infty$ and for the one-sided limit as $x \to c+$.  The cases where $f'(x)/g'(x) \to -\infty$  and $x \to c-$ are proved with some easy modifications.
(1) Proof with the hypothesis $\displaystyle \lim_{x \to c+} f(x) = \lim_{x \to c+} g(x) = 0$.
For every $K > 0$ there exists $\delta_k > 0$ such that if $c < x < c + \delta_K$ then $\frac{f'(x)}{g'(x)} > K$.
Take any $y$ such that $c < y < x < c +\delta_K$. By Cauchy's mean value theorem there exists $\xi \in (y,x)$ such that
$$\frac{f(x) - f(y)}{g(x) - g(y)} = \frac{f'(\xi)}{g'(\xi)} > K,$$
where the inequality on the RHS follows because $c < y < \xi < x < c+\delta_K$.  Taking the limit as $y \to c+$ we get for all $ c < x < c+ \delta_K$
$$\frac{f(x)}{g(x)} = \lim_{y \to c+} \frac{f(x) - f(y)}{g(x) - g(y)} \geqslant K,$$
and this implies the desired result.
(2) Proof with the hypothesis $\displaystyle \lim_{x \to c+} f(x) = \lim_{x \to c+} g(x) = +\infty$.
For every $K > 0$ there exists $\delta_k > 0$ such that if $c < x < c + \delta_K$ then $\frac{f'(x)}{g'(x)} > \frac{3K}{2}$.
Take any $y$ such that $c < x < y < c +\delta_K$. By Cauchy's mean value theorem there exists $\xi \in (y,x)$ such that
$$\frac{f(x)}{g(x)}\frac{1 - \frac{f(y)}{f(x)}}{1 - \frac{g(y)}{g(x)}} = \frac{f(x) - f(y)}{g(x) - g(y)} = \frac{f'(\xi)}{g'(\xi)} > \frac{3K}{2},$$
where the inequality on the RHS follows because $c < y < \xi < x < c+\delta_K$.  Since $1/f(x) \to 0$ and $1/g(x) \to 0$ as $x \to c+$, we have
$$\lim_{x \to c+} \frac{1 - \frac{f(y)}{f(x)}}{1 - \frac{g(y)}{g(x)}} = 1$$
There exists $\delta_1 > 0$ such that if $c < x < c +\delta_1$, then
$$\frac{1}{2} < \frac{1 - \frac{f(y)}{f(x)}}{1 - \frac{g(y)}{g(x)}} < \frac{3}{2}$$
Thus, if $c < x < c + \min(\delta_K, \delta_1)$, we have
$$\frac{3}{2}\frac{f(x)}{g(x)} > \frac{f(x)}{g(x)}\frac{1 - \frac{f(y)}{f(x)}}{1 - \frac{g(y)}{g(x)}} = \frac{f(x) - f(y)}{g(x) - g(y)} = \frac{f'(\xi)}{g'(\xi)} > \frac{3K}{2},$$
and this implies the desired result.
