It seems that when the logarithmic derivative of a function exists, the function itself cannot be zero. But I failed to construct a convincent proof. Is this true?
$\dfrac{f'(x)}{f(x)}$ is bounded implies $f(x)$ cannot have zero as a limiting value.
Is this true?
Let $f:x\to \frac{1}{x}$ on $[1,+\infty)$, then we have $$\frac{f'(x)}{f(x)}=-\frac{1}{x},$$ so $\displaystyle\frac{f'}{f}$ is bounded on $[1,+\infty)$ Nevertheless $\displaystyle \lim_{+\infty}f=0.$
True when $x$ approaches a finite $x_0$, not when $x$ approaches $\pm \infty$, as explained below.
Assume $f$ is differentiable on $\mathbb{R}$.
Then the function $$ g(x):=\ln|f(x)| $$ is defined and differentiable on the open set $$ U=\{x\in\mathbb{R}\;;\; f(x)\neq 0\}. $$ Its derivative is $$ g'(x)=\frac{f'(x)}{f(x)}\qquad \forall\;x\in U. $$
Write $U=\bigcup_{n\geq 1}(a_n,b_n)$ as a countable disjoint union of open intervals. See this thread to see why this is possible.
Consider for instance a finite point $x_0=a_n$.
Note that $\lim_{x\rightarrow x_0^+}\ln |f(x)|=\lim_{x\rightarrow x_0^+}g(x)=-\infty$.
Now if $g'=(\ln |f|)'$ was bounded by, say, $M$ on $(x_0,x_0+\delta]$, the mean value theorem, and then the triangular inequality, would yield $$ |g(x)-g(x_0+\delta)|\leq M \delta\quad \Rightarrow\quad |g(x)|\leq |g(x_0+\delta)|+M\delta $$ for all $x\in(x_0,x_0+\delta]$.
Contradiction.
So the logarithmic derivative is unbounded when it approaches every finite boundary point of $U$.
Now regarding the behavior at $\pm\infty$, you can easily find examples where $f'/f$ is bounded while $f$ tends to $0$.
