limit of $\lim_{x\to\infty} \frac{f(x)}{f'(x)} $

Question

Suppose there exists a function $ f(x) \: s.t. \: f(x),{f}'(x),{f}''(x) >0 \: \forall x\in \mathbb{R} \\ $

We can conlude from the above that $f(x)$ is strictly convex and $ \lim_{x\to\infty} f(x) =\infty \: \text{and} \: \lim_{x\to\infty} f'(x) =\infty .$

However, how may one prove that $ \lim_{x\to\infty} \frac{f(x)}{f'(x)}\geq 1 ? $

Answer 1

You made an edit to the question, but the claim still appears wrong. Consider: $$ f(x) = e^{\lambda x} $$ Then, $$ \frac{f(x)}{f'(x)} = \frac{1}{\lambda} $$

Also, you claim that $f, f', f'' > 0$ implies that $f(x) \to \infty$ and $f'(x) \to \infty$ as $x \to \infty$. But this isn't true either. Take $f(x) = \log( e^x + 1)$, $f'(x) = \frac{1}{1+e^{-x}}$, and $f''(x) = \frac{e^{-x}}{(1+e^{-x})^2}$. In this case, $\lim\limits_{x \to \infty} f'(x) = 1$.

Answer 2

One can't. Take $f(x)=e^{2x}$. Then$$\lim_{x\to\infty}\frac{f(x)}{f'x)}=\frac12.$$