Forum for the GRE subject test in mathematics.
speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

Hey guys, Look at this limit
$$\lim_{x\rightarrow0}\frac{1}{x^2}\int_{0}^{x}\frac{t+t^2}{1+sint}d t$$

If I use L'Hopital's rule directly, I can get the limit is 1/2

but as I tried to solve the problem in another approach, the question occured,

let$$f(t)=\frac{t+t^2}{1+sint}$$ and $$F(t)=\int\frac{t+t^2}{1+sint}d t$$

then the limit could be written as

$$\lim_{x\rightarrow0}\frac{F(x)-F(0)}{x^2}$$ and since $$x\rightarrow0$$

the term could then be written as
$$\lim_{x\rightarrow0}\frac{f(0)}{x}$$
and now use the L'Hopital' rule,
we will get the limit is 1

I know there is something wrong, but would you mind pointing out where I have made mistakes? thanks.
Last edited by speedychaos4 on Wed Apr 07, 2010 12:56 am, edited 1 time in total.

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

### Re: Doubts about L'Hopital's Rule

I'm not following, your logic is sort of obscure (no offense).
first, you write 't' in the end of integrand, while it should be 'dt', but that's minor.
what is more significant is that you write f(x), but your right-hand side depends only on variable t, and the same is for F(x).

p.s. there's a very neat way how to solve this [easy] problem even w/o L'Hopital.
notice that you work with very little x (x -> 0), and recall that for such x "sin(x) ~= x" holds true. thus, the initial integral becomes
$$\lim\limits_{x\rightarrow0}\frac1{x^2}\int\limits_{0}^{x}\frac{t+t^2}{1+t}\,dt = \lim\limits_{x\rightarrow0}\frac1{x^2} \int\limits_{0}^{x} t \,dt = \lim\limits_{x\rightarrow0}\frac1{x^2} \frac{x^2}{2} = \frac12.$$

enork
Posts: 33
Joined: Fri Sep 18, 2009 3:16 am

### Re: Doubts about L'Hopital's Rule

You replace $$\frac{F(x)-F(0)}{x}$$ with f(0) but that's not valid. $$\lim_{x \to 0}\frac{F(x)-F(0)}{x} = f(0)$$ but that's not what you have.

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: Doubts about L'Hopital's Rule

EugeneKudashev wrote:I'm not following, your logic is sort of obscure (no offense).
first, you write 't' in the end of integrand, while it should be 'dt', but that's minor.
what is more significant is that you write f(x), but your right-hand side depends only on variable t, and the same is for F(x).

p.s. there's a very neat way how to solve this [easy] problem even w/o L'Hopital.
notice that you work with very little x (x -> 0), and recall that for such x "sin(x) ~= x" holds true. thus, the initial integral becomes
$$\lim\limits_{x\rightarrow0}\frac1{x^2}\int\limits_{0}^{x}\frac{t+t^2}{1+t}\,dt = \lim\limits_{x\rightarrow0}\frac1{x^2} \int\limits_{0}^{x} t \,dt = \lim\limits_{x\rightarrow0}\frac1{x^2} \frac{x^2}{2} = \frac12.$$
sorry for the mistakes, I have corrected now. Due to enork's explanation, I understand what is my problem, I'm messing up the property of lim.

and you "sinx~=x" approach really shed the lights on!! Thanks!!

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: Doubts about L'Hopital's Rule

enork wrote:You replace $$\frac{F(x)-F(0)}{x}$$ with f(0) but that's not valid. $$\lim_{x \to 0}\frac{F(x)-F(0)}{x} = f(0)$$ but that's not what you have.
Thanks, I kinda incorrectly assume that limab=limalimb

origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

### Re: Doubts about L'Hopital's Rule

speedychaos4 wrote:Thanks, I kinda incorrectly assume that limab=limalimb
It generally does when those limits exist, but in this case they don't ($$\lim_{x \rightarrow 0} \frac{1}{x}$$)