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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!

Arctic Char
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Lee Laindingold
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  • Yes, if the limit of the quotient of the derivatives exists, finite or infinite, then it is the same as the quotient of the functions. – egreg Oct 16 '20 at 07:51
  • Is there a proof for when the limit of the quotient of the derivatives is infinite? Btw, the limit exists actually implies that the limit is not infinity-- it has to be equal to a number, is that correct? – Lee Laindingold Oct 16 '20 at 10:37
  • You find proofs of the theorem in many books. What “the limit exists” means depends on the conventions set out by your book or instructor. – egreg Oct 16 '20 at 10:56
  • The proofs in the book i have all seem to agree on L'Hospital's Rule only when the limit of quotient of derivatives is equal to a number-- could you provide a proof where the limit tends to inifinity? Or a link would be nice! Thanks! – Lee Laindingold Oct 17 '20 at 00:48
  • You can reduce the proof to the limit being $0$ and considerations on the signs, by using the reciprocal. – egreg Oct 17 '20 at 09:06

1 Answers1

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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.

RRL
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