$f(x)\in C^0,f(0)=0,$ for$\ x\neq 0,f(x)>0,$ is $\int^x_0f(t)dt=o(f(x))?$ I wanna find a counter example to prove that "if $f(x)\in C^0,f(0)=0$,and for $x\neq 0(f(x)>0)$, then $\int^x_0f(t)dt=o(f(x)),$ as $x\to 0$" is not always true even if $f(x)\in C^p,p\geq1$.
First I found $f(x)=x^n|\sin(\frac1x)|+x^m\in C^0,m>n$, and by using an auxiliary function $\phi(x)$ which is a broken line connecting the point where $|sin(\frac1x)|=0,1$. So for the sequence $\{x_k=\frac1{k\pi}\},\int_0^{x_k}t^n\sin(\frac1t)dt\geq\int_0^{x_k}t^n\sin\phi(t)dt\geq\frac{nx_k^{n+1}}{2(n+1)}-\frac{(x_k)^{n+1}}{n\pi}$ , which means $\lim\limits_{k\to\infty} \frac{\int^{x_k}_0f(t)dt}{f(x)}$ doesn't exist.
Then I found that $f(x)=x^n(\sin(\frac1x)+1)+x^m\in C^{n-2},m>n$, and by using $\varphi(x)=$$\begin{cases}1,sin(\frac1x)>0\\0,sin(\frac1x)\leq0\end{cases}$ and $\int^{x_k}_0t^n\varphi(x)dt\geq\int^{\frac{x_k}2}_0t^ndt$, we can prove that for  $\{x_k=\frac{1}{2k\pi-\frac{\pi}2}\}, \lim\limits_{k\to\infty}\frac{\int^{x_k}_0f(t)dt}{f(x)}$ doesn't exist.
As you can see, these two functions are pretty complicated, so I'm wondering if there are better (simpler) counter examples for the problem.
 A: My counterexample:
$$
f(x) = \exp\left(\frac{-1}{x}\left(2+\sin\frac{1}{x}\right)
\right),\qquad x>0
$$
Then $f(x) \to 0$ as $x \to 0^+$, but $\int_0^x f(t)\;dt$ is not $o(f(x))$.

How did I find this?
First, $\int_0^x f(t)\;dt = o(f(x))$ as $x \to 0^+$ means
$$
\lim_{x\to 0^+} \frac{\int_0^x f(t)\;dt}{f(x)} = 0
$$
Note $f(x) \to 0$ and $\int_0^x f(t)\;dt \to 0$, so by l'Hopital we should attempt
the limit of
$$
\frac{\frac{d}{dx}\int_0^x f(t)\;dt}{f'(x)} = \frac{f(x)}{f'(x)}
= \frac{1}{\frac{d}{dx}\log(f(x))}
\tag1$$
For our counterexample, we want $\frac{d}{dx}\log(f(x))$ does not go to $\pm \infty$.  We do have $f(x) \to 0$, so $\log(f(x)) \to -\infty$.  So
I found a function that goes to $-\infty$, but oscillates so that its derivative does not go to $\infty$.  I used
$$
g(x) := \frac{-1}{x}\left(2+\sin\frac{1}{x}\right)
$$
with graph like this:


Computation.  Define $x_n = \frac{1}{(2 n-1) \pi}$, so that $x_n \to 0^+$,
and define interval
$$
J_n = \left[\frac{1}{(2n-\frac16)\pi},\frac{1}{(2n-\frac56)\pi}\right]
$$
so that $\sin\frac1x \le -\frac12$ for $x \in J_n$.
Then
$$
\int_0^{x_n} f(x)\;dx > \int_{J_n} f(x)\;dx >
 \frac{24}{(12n-1)(12n-5)\pi}\;\exp\left(-\frac32 (2n-1)\pi\right)
$$
and
$$
f(x_n)  = \exp\big(-(2n-1)\pi\cdot (2+0)\big)
$$
Divide
$$
\frac{1}{f(x_n)}\int_0^{x_n} f(t)\;dt >
\frac{24}{(12n-1)(12n-5)\pi}\;\exp\left(\pi n - \frac12 \pi\right)
\to \infty \text{ as }n \to \infty .
$$
So
$$
\frac{1}{f(x)}\int_0^{x} f(t)\;dt \not\to 0\quad\text{ as } x \to 0^+ .
$$
A: Many counterexamples can be obtained this way: Let $(x_n)$ be a sequence in $(0,1]$ that decreases to $0.$ Let $g$ be a continuous function on $[0,1]$ such that $g(x)=0$ for all $x\in \{x_1,x_2,\dots\}\cup \{0\},$ and $g>0$ elsewhere. (Example: $g(x)=x\sin^2(1/x),$ $x_n=1/(n\pi), n=1,2,\dots$)
Define
$$f(x) = g(x) + \left(\int_0^x g(t) \,dt\right)^2.$$
This $f$ will be a counterexample. Note $f(x_n) = (\int_0^{x_n} g(t) \,dt)^2$ for all $n.$ Thus as $n\to \infty,$
$$\frac{\int_0^{x_n} f(t)\,dt}{f(x_n)} > \frac{\int_0^{x_n} g(t)\,dt}{(\int_0^{x_n} g(t) \,dt)^2} = \frac{1}{\int_0^{x_n} g(t) \,dt} \to \infty.$$
