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Let $f(x)$ be a function, the second derivative $f^{\prime\prime}(x)$ exists, and $\lim\limits_{x\to +\infty}\dfrac{f(x)}{e^x}=l$. $c\gt0$ such that for sufficiently large $x$, $|f''(x)|<c|f'(x)|$. then we have $$\lim_{x\to + \infty}\frac{f'(x)}{e^x}=l $$

converse of L'Hôpital's rule ?

Thank you very much for your help

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  • $\begingroup$ And what's the question exactly? $\endgroup$ – Daniel Robert-Nicoud Jun 3 '14 at 19:05
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I think I got it :

lets define $G(x) = f(x)/e^{x} $ we have $G'(x) = \frac {f'(x)}{e^{x}} - \frac {f(x)}{e^{x}}$

  • for $l < \infty $ :

Suppose $ \lim_{x\to + \infty}G'(x) \neq 0 $

this means that $ \exists\ \epsilon > 0 $ for which we can find a monotone sequence ($x_{n}) $ validating : $ |G(x_{n})| > \epsilon $

The mean value theorem allows use to find a sequnce $ c_{n} \in [x_{n},x_{n}+1] $ so that :

$$ G(x_{n}+1) - G(x_{n}) = G'(c_{n})(x_{n}+1-x_{n}) = G'(c_{n}) $$

Which gives us :

$$ G(x_{n}+1) - G(x_{n}) = \frac {f'(c_{n})}{e^{c_{n}}} - \frac {f(c_{n})}{e^{c_{n}}} $$

The first par tends toward $0$, so $\frac {f'(c_{n})}{e^{c_{n}}} \sim l$.

The inequality $ |f''(x)|<c|f'(x)| $ and the mean theorem allows use to say that $ f' $ is continue.

since $c_{n}\sim x_{n}$, $G(x_{n}) \sim 0 $. Absurde.

  • for $l = +\infty $ :

Suppose $\lim_{x\to + \infty}\frac {f'(x)}{e^{x}} \neq +\infty$.

$ \exists\ M > 0 $ for which we can find a monotone sequence ($x_{n}) $ so that : $ \frac {f'(x_{n})}{e^{x_{n}}} < M $.

Thus $ G'(x_{n}) = \frac {f'(x_{n})}{e^{x_{n}}} - \frac {f(x_{n})}{e^{x_{n}}} \sim -\infty $. Starting some range $m $, $ n \geq m \Rightarrow G'(x_{n}) < 0 $ (because of the continuity).

The mean theorem again, $ c_{n} \in [x_{m},x_{n}] $ so that :

$$ G(x_{n}) - G(x_{m}) = G'(c_{n})(x_{n}-x_{m}) $$ and $$ G'(c_{n}) < 0 $$

We tend n to $+\infty$, wich gives : $G(x_{n})$ tends to $-\infty$. Absurde.

  • For $l = -\infty$ : Take $f = -f$.

I hope it's correct.

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  • $\begingroup$ No, it's not correct $\endgroup$ – ziang chen Jun 4 '14 at 23:08
  • $\begingroup$ You assume that either $G'(x)$ tends to a limit or to $\infty$ but it may happen that it oscillates. What you have done is a simple theorem : if $G(x) \to L$ as $x \to \infty $ and $G'(x)$ also tends to a limit, say $L'$ then $L'=0$. $\endgroup$ – Paramanand Singh Jun 5 '14 at 3:06
  • $\begingroup$ It's not $G'(x)$ that has a limit, but $G'(x_{n})$, and $x_{n}$ is constructed from the definition of non convergence. I did not use "$G'(x)$ tends to a limit". Or did I ? $\endgroup$ – Naoufal EL JAOUHARI Jun 5 '14 at 12:37
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    $\begingroup$ Well if you carefully note it what you have actually shown is that there is a sequence $c_{n}$ such that $c_{n} \in (x_{n}, x_{n} + 1)$ and $G'(c_{n}) \to 0$ and this is not incompatible with $|G'(x_{n})| > \epsilon$. If $f$ is any differentiable (or say infinitely differentiable) then there can be sequences $x_{n}, y_{n}$ such that $f(x_{n})$ tends to a limit and $f(y_{n})$ does not tend to a limit. In your we have $G'(c_{n}) \to 0$ and $G'(x_{n})$ not tending to $0$. $\endgroup$ – Paramanand Singh Jun 7 '14 at 4:52
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    $\begingroup$ I believe one needs to use the second condition $|f''(x)| < c|f'(x)|$ to specify more control on the behavior of $G'(x)$ and then perhaps we can achieve something. $\endgroup$ – Paramanand Singh Jun 7 '14 at 5:02
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Note that $\lim_{x \rightarrow +\infty} \frac{f(x+1) - f(x)}{e^{x+1} - e^x}=l$ since

$$ \frac{f(x+1) - f(x)}{e^{x+1} - e^x}=\frac{1}{e^{x+1}} \frac{f(x+1) - f(x)}{1 - e^{-1}} $$ $$= \left[ \frac{f(x+1)}{e^{x+1}} - \frac{1}{e} \frac{f(x)}{e^x} \right] \left( \frac{1}{1-e^{-1}} \right) $$ $$\rightarrow \left[ l - \frac{l}{e} \right]\left( \frac{1}{1 - e^{-1}} \right), \ x \rightarrow +\infty$$ $$=l$$

By MVT, there is a $x < y < x + 1$ such that $\frac{f(x+1) - f(x)}{e^{x+1} - e^x}=\frac{f'(y)}{e^y}$.

Suppose $c \le 1$. By hypothesis, there is an $A>0$ such that $|f'(x)|>|f''(x)|$ if $x > A$. In particular, $f'$ has no zeros for $x>A$, so let's suppose $f'>0$.

Let $h(x)=\frac{f'(x)}{e^{x}}$. Since $\left| \frac{d}{dx} \log(f'(x)) \right| = \left| \frac{f''(x)}{f'(x) } \right| < 1$, we have that $\log(f'(x)) - 1 = \log h(x)$ is Lipschitz with constant $1$, by the MVT. It follows that $h$ is decreasing and positive, so $\lim_{x \rightarrow +\infty} h(x)$ exists.

Given a positive integer $n$, take $n < y_n < n + 1$ such that $\frac{f(n+1) - f(n)}{e^{n+1} - e^n}=h(y_n)$.

Therefore, $\lim_{x \rightarrow +\infty}\frac{f'(x)}{e^x}=\lim h(y_n)=l$.

For $c>1$, apply the result for $g(x)=f(\frac{x}{c})$. (Not so sure about this piece.)

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