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.
Doubts about L'Hopital's Rule

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Doubts about L'Hopital's Rule
Last edited by speedychaos4 on Wed Apr 07, 2010 12:56 am, edited 1 time in total.

 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 righthand side depends only on variable t, and the same is for F(x).
try again, please.
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
$$\lim\limits_{x\rightarrow0}\frac1{x^2}\int\limits_{0}^{x}\frac{t+t^2}{1+t}\,dt = \lim\limits_{x\rightarrow0}\frac1{x^2} \int\limits_{0}^{x} t \,dt = \lim\limits_{x\rightarrow0}\frac1{x^2} \frac{x^2}{2} = \frac12.$$
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 righthand side depends only on variable t, and the same is for F(x).
try again, please.
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
$$\lim\limits_{x\rightarrow0}\frac1{x^2}\int\limits_{0}^{x}\frac{t+t^2}{1+t}\,dt = \lim\limits_{x\rightarrow0}\frac1{x^2} \int\limits_{0}^{x} t \,dt = \lim\limits_{x\rightarrow0}\frac1{x^2} \frac{x^2}{2} = \frac12.$$
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.

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: Doubts about L'Hopital's Rule
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.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 righthand side depends only on variable t, and the same is for F(x).
try again, please.
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
$$\lim\limits_{x\rightarrow0}\frac1{x^2}\int\limits_{0}^{x}\frac{t+t^2}{1+t}\,dt = \lim\limits_{x\rightarrow0}\frac1{x^2} \int\limits_{0}^{x} t \,dt = \lim\limits_{x\rightarrow0}\frac1{x^2} \frac{x^2}{2} = \frac12.$$
and you "sinx~=x" approach really shed the lights on!! Thanks!!

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: Doubts about L'Hopital's Rule
Thanks, I kinda incorrectly assume that limab=limalimbenork 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.
Re: Doubts about L'Hopital's Rule
It generally does when those limits exist, but in this case they don't ($$\lim_{x \rightarrow 0} \frac{1}{x}$$)speedychaos4 wrote:Thanks, I kinda incorrectly assume that limab=limalimb