if $f(x)=o(g(x))$ will $\int_0^x f(x)=o(\int_0^x g(x))$ and $f'(x)=o(g'(x))$? Let $f$ and $g$ be two functions with derivatives in some interval containing $0$, where $g$ is positive. Also
$$f(x)=o(g(x))~as~x \rightarrow0$$
Prove or dissprove:
1) $$\int_0^xf(t)dt=o\left(\int_0^xg(t)dt\right)$$
2) $$f'(x)=o(g'(x))$$
Now considering the first, my reasoning is as follows:
$$\lim_{x\to0}\frac{\int_0^xf(t)dt}{\int_0^xg(t)dt}=\lim_{x\to0}\frac{xf(x)-\int_0^xtf'(t)dt}{\int_0^xg(t)dt}$$
Now the first member on the right will tend to $0$. Second will also seems to converge to $0$ (though I am unsure of that). And the limit should converge to $0$? I realize this is a very weak reasoning. How could I make it more precise?
Considering the second problem I am quite clueless though I am quite sure it should converge to $0$ :) Any hints?
 A: 1) Begin by writing down the meaning of $o$: for any $\epsilon>0$ there is $\delta>0$ such tha t $$|f(x)|\le \epsilon \, g(x) \quad \text{for all }\ x\in (-\delta,\delta) \tag1$$ 
Then use the integral triangle inequality: 
$$\left|\int_0^x f(t)\,dt\right|\le \int_0^x |f(t)|\,dt \le \epsilon\,\int_0^x g(t)\,dt$$
2) Already answered by Steven Stadnicki: the functions $f(x)=\sin x$ and $g(x)=x+1$ satisfy $f=o(g)$ but fail $f'=o(g')$.
A: Considering the second - I can not use the L'Hopital's rule yet, so I have to stick to other means. I think I found away, but I need help confirming weather such approach is viable:
Let $F(x)=\int_0^xf(t)dt$ and $G(x)=\int_0^xg(t)dt$. By definition:
$$\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=F'(x)+o(1)$$
$$\lim_{h\to 0}\frac{G(x+h)-G(x)}{h}=G'(x)+o(1)$$
So since
$$0=\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to 0}\frac{F'(x)+o(1)}{G'(x)+o(1)}$$
It follows that $$\int_0^xf(t)dt=o\left(\int_0^xg(t)dt\right)$$
Is this correct?
For the first - Steven's example is enough to disprove my "quite sure" position. But how could I make it more formal? Can not find any ideas.
