# If $\lim_{x\to\infty}\frac{f'(x)}{f(x)}=1$, is $f$ asymptotic to $\exp$?

For a differentiable and nonzero function $$f:(a,\infty)\to\mathbb R$$, it seems like the local and end behavior of $$f'(x)/f(x)$$ gives a measure for how similar $$f$$ is to an exponential function. This is a reasonable guess because the quantity $$f'(x)/f(x)$$ gives the derivative of $$\ln\circ f$$ at $$x$$. My suspicion began when I noticed that for any exponential $$a^x$$, we have

$$\frac{f'(x)}{f(x)}=\frac{a^x\ln(a)}{a^x}=\ln(a)$$

indicating that exponential functions are the only functions for which $$f'/f$$ is constant. Exploring this further with the hyperbolic cosine $$\cosh(x)$$, a function which is basically identical to $$\exp(x)$$ as $$x\to\infty$$, I arrived at

$$\frac{f'(x)}{f(x)}=\frac{\sinh(x)}{\cosh(x)}=\tanh(x)\to 1\text{ as }x\to\infty$$

These examples were enough to convince me, perhaps erroneously, that $$f'/f$$ gives a measure for how similar a function is to an exponential, leading me to the following conjecture:

Suppose $$f:(a,\infty)\to\mathbb R$$ is a differentiable function that is never zero. If $$\lim_{x\to\infty}\frac{f'(x)}{f(x)}=L$$, does it follow that for some constant $$C$$, $$f(x)\sim C\exp(Lx)$$ as $$x\to\infty$$?

For simplicity, I focused my attention to the case where $$f'(x)/f(x)\to 1$$, $$f$$ is strictly positive, and $$f'/f$$ is "eventually integrable", i.e. there is a constant $$c$$ such that for every $$x$$ greater than $$c$$, $$f'/f$$ is integrable over $$[c,x]$$. Unraveling $$\lim_{x\to\infty}\frac{f'(x)}{f(x)}=1$$ with the definition of a limit and leveraging these simplifying assumptions, I was able to deduce the following:

For every $$\varepsilon>0$$, there is a $$\delta\in\mathbb R$$ such that for some constants $$C_1,C_2>0$$, $$C_1\exp\left((1-\varepsilon)x\right)\delta$$

As we make $$\varepsilon$$ smaller and smaller, the functions $$\exp\left((1-\varepsilon)x\right)$$ and $$\exp\left((1+\varepsilon)x\right)$$ will look more and more like $$\exp(x)$$, suggesting that the ultimate goal of $$\lim_{x\to\infty}\frac{f(x)}{C\exp(x)}=1$$ might be true. However, it doesn't seem like my result is sufficient to arrive at this conclusion. The constants $$C_1,C_2>0$$ depend on $$\varepsilon$$ and $$\delta$$, so I'm afraid they may fluctuate enough to ruin any hopes of asymptotic equivalence.

Is my simplified conjecture, equipped with the integrability assumption, actually true? If so, how can I reach the goal? If not, what other assumptions on $$f$$ are needed to ensure asymptotic equivalence?

• How do you define asymptotic equivalence? Commented Jan 11, 2022 at 6:35
• @Raskolnikov $f(x)\sim g(x)$ as $x\to\infty$ if and only if $\lim_{x\to\infty}\frac{f(x)}{g(x)}=1$. Commented Jan 11, 2022 at 6:36
• $f(x)=x e^x$ satisfies your limit property. Commented Jan 11, 2022 at 6:36
• $\log f$ is asymptotic to $x$ but this doesn't imply $f\sim \exp(x)$ Commented Jan 11, 2022 at 7:05

No. If we set $$f(x) = e^{g(x)}$$, then $$\frac{f'(x)}{f(x)} = g'(x)$$, and so the question is asking whether $$\lim_{x\to\infty} g'(x) = 1$$ implies that $$g(x)$$ is linear. But there are plenty of functions $$h(x)$$ such that $$\lim_{x\to\infty} h(x)=1$$ yet $$h(x)$$ is not identically $$1$$, and we can simply let $$g(x)$$ be an antiderivative of any such $$h(x)$$. For example, setting $$h(x) = 1+\frac1x$$ yields $$g(x) = x+\ln x$$ and so $$f(x) = xe^x$$, as commented by David Mitra.

• So, what we do know: $e^{bx} < f(x) < e^{ax}$ for large $x$ for every $a,b$ with $b<1<a$. Commented Jan 11, 2022 at 12:16

Let $$f(x)=xe^x$$. Then $$\frac{f'(x)}{f(x)}=\frac{xe^x+e^x}{xe^x}=1+\frac{1}{x}\xrightarrow{x\to\infty}1,$$ but $$f$$ is not asymptotic to an exponential function.

The problem is exactly what you state; $$C_1$$ and $$C_2$$ depend on $$\epsilon$$.

Let $$g(x)=f(x)e^{-x}.$$ Then $$\frac{g’(x)}{g(x)}=\frac{f’(x)e^{-x}-f(x)e^{-x}}{f(x)e^{-x}}=\frac{f’(x)}{f(x)}-1\to 0.$$

So you can investigate what $$g(x)$$ satisfy $$g’(x)/g(x)\to 0$$. But $$\frac{g’(x)}{g(x)}=\left(\ln |g(x)|\right)'$$

So when $$h(x)=\ln |g(x)|$$ then you need $$h’(x)\to 0.$$

Given such an $$h,$$ you can define $$g(x)=\pm e^{h(x)}.$$ Then you get $$f(x)=\pm e^{x+h(x)}.$$ Then $$f’(x)=f(x)(1+h’(x)).$$

That gives you all such continuous $$f.$$

This means that $$f(x)e^{-x}$$ can be fairly huge. For example, $$f(x)=e^{x+\sqrt x}$$ works. The best we can say is $$\log|f(x)|\sim x.$$

If $$L \neq 0$$ : $$\dfrac{f'(x)}{f(x)} \underset{+\infty}{\sim} L$$ gives : $$\int_a^x \dfrac{f'(t)}{f(t)} \mathrm{d}t \underset{+\infty}{\sim} \int_a^x L \mathrm{d}t$$ then : $$\ln |f(x)| \underset{+\infty}{\sim} L x$$ So : $$f(x) \underset{+\infty}{\sim} e^{L x} e^{o(x)}$$