integral solution

[Vali]

\(\displaystyle \lim_{n\rightarrow \infty }\frac{1}{n}\int_{1}^{n}\frac{x-1}{x+1}dx\)
My approach is not correct, I think.
I took f(x)=(x-1)/(x+1) which is continuous so there is a F(x) a primitive of f(x) such that F'(x)=f(x)
So I used L'Hospital but when I derived, I derived for n which is number, not for x and I think it's not correct.
I obtained \(\displaystyle \lim_{n \to \infty }\frac{F(n)-F(1)}{n}=\lim_{n \to \infty }f(n)=1\) which is the correct answer but I'm not sure if this is the right method.There's another way to solve this?

 

[pka]

\(\displaystyle \frac{d}{{dn}}\left( {\int_1^n {f(t)dt} } \right) = f(n)\)

 

[Steven G]

Usually this is another method. You can write (x-1)/(x+1) as ((x+1)-2)/(x+1) = 1 - 2/(x+1). This integral will then be x-2ln(x+1)......

 

[MarkFL]

I think the key here is to observe that the integral with the upper bound unbounded does not converge, and so L'Hôpital's Rule may be applied. I would write:

[MATH]L=\lim_{n\to\infty}\left(\frac{\displaystyle\int_1^n \dfrac{x-1}{x+1}\,dx}{n}\right)=\lim_{n\to\infty}\left(\frac{n-1}{n+1}\right)=1[/MATH]
Of course we could evaluate the integral too:

[MATH]\int_1^n \frac{x-1}{x+1}\,dx=\int_1^n 1-\frac{2}{x+1}\,dx=\left[x-2\ln(x+1)\right]_1^n=n-2\ln(n+1)-1+\ln(2)[/MATH]
From this, we may conclude:

[MATH]L=1-2\lim_{n\to\infty}\left(\frac{\ln(n+1)}{n}\right)=1[/MATH]

 

[Vali]

Thank you for your answers!