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 $C_1,C_2>0$,
$$C_1\exp\left((1-\varepsilon)x\right)<f(x)<C_2\exp\left((1+\varepsilon)x\right)\text{ for every }x\in\text{dom}[f'/f]\text{ with }x>\delta$$

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?
 A: 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 $\epsilon$.
A: 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 $$\log|f(x)|\sim x.$$
A: 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 :
$$f(x) \underset{+\infty}{\sim} e^{L x} e^{o(x)}$$
A: 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.
