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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?

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1  
For 2: Try $f(x) = \sin(x), g(x) = x$ (or if you prefer, $g(x) = x+1$ since it's positive on an interval containing 0 - that's a detail here.)? The point is that knowing bounds on something doesn't give you any bounds on how much it can 'wiggle'. –  Steven Stadnicki Jan 10 '13 at 22:17
    
"Now considering hte first, my reasoning is as follows:..." Apply L'Hopital's rule for an immediate answer. –  Alex R. Jan 11 '13 at 1:21
    
You are doing fine with the first one, but you should apply L'hopital's rule correctly. You should have $ \lim_{x\to0}\frac{\int_0^xf(t)dt}{\int_0^xg(t)dt}= \lim_{x\to0}\frac{f(x)}{g(x)}=0 $. –  Mhenni Benghorbal Jun 25 '13 at 12:20

2 Answers 2

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 $f'=o(g')$.

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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.

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so could anybody verify my reasoning and/or provide any hints regarding other part of the problem? thanks! –  Sarunas Jan 13 '13 at 21:25
    
I do not understand the role of $o(1)$ in $\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=F'(x)+o(1)$. By definition of derivative, $\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=F'(x)$. Also, "it follows that" appears to be based on the L'Hopital's rule which you said you can't use. –  ˈjuː.zɚ79365 Jun 25 '13 at 12:01

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