# Finding $lim_{n to infty} int_{2}^{infty} frac{nsinleft(frac{x-2}{n}right)}{(x-2)+(1+(x-2)^2)} dx$

Mathematics Asked by User160 on February 21, 2021

I have to calculate the following limits, using a theorem but I don’t really know what theorem to use (it is for the subject of measurement and integration, for the unit "Measurable functions, integration and its properties"). $$space$$

$$lim_{n to infty} int_{2}^{infty} frac{nsinleft(frac{x-2}{n}right)}{(x-2)+(1+(x-2)^2)} dx$$

Do I have to use the dominated convergence theorem of Lebesgue?

I have first of all calculated $$limlimits_{n to infty}frac{nsinleft(frac{x-2}{n}right)}{(x-2)+(1+(x-2)^2)}$$ and I’ve obtained $$frac{1}{2x-3}$$

Now, I want to calculate $$int_{2}^{infty} frac{1}{2x-3},dx$$ but $$ln(infty)$$ doesn’t exist… so what am I doing wrong?

I would "drop the tail" and use the monotone convergence theorem (which doesn't require the finiteness).

Our integral is $$I_n=int_0^inftyfrac{nsin(x/n)}{1+x+x^2},dx=J_n+R_n$$ with $$J_n=int_0^{npi}$$ and $$R_n=int_{npi}^infty$$. Now $$|R_n|leqslant nint_{npi}^inftyfrac{dx}{x^2}=frac1pi$$, i.e. $$R_n$$ is bounded, and $$limlimits_{ntoinfty}J_n=+infty$$ by MCT (indeed, $$J_n=int_0^infty f_n(x),dx$$ with $$f_n(x)=0$$ for $$xgeqslant npi$$, and $$nmapsto f_n(x)$$ is nondecreasing for each fixed $$x$$, because $$tmapsto(sin t)/t$$ is decreasing for $$tin[0,pi]$$). Thus, $$limlimits_{ntoinfty}I_n=+infty$$.

Answered by metamorphy on February 21, 2021