# $\lim_{x\to+\infty}\frac{f(x)}{e^x}=l$, then $\lim_{x\to+\infty}\frac{f'(x)}{e^x}=l$?

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 $$x$$, $$|f''(x)|. then we have $$\lim_{x\to + \infty}\frac{f'(x)}{e^x}=l$$

converse of L'Hôpital's rule ?

Thank you very much for your help

• And what's the question exactly? – Daniel Robert-Nicoud Jun 3 '14 at 19:05

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.

• No, it's not correct – ziang chen Jun 4 '14 at 23:08
• 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 $L'=0$. – Paramanand Singh Jun 5 '14 at 3:06
• 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 ? – Naoufal EL JAOUHARI Jun 5 '14 at 12:37
• 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 $0$. – Paramanand Singh Jun 7 '14 at 4:52
• 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. – Paramanand Singh Jun 7 '14 at 5:02

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.)