Matematické Fórum

stuart clark · 22. 12. 2018 09:02

If $a_{n}=\int^{\frac{1}{n}}_{\frac{1}{n+1}}\tan^{-1}(nx)dx$ and $b_{n}=\int^{\frac{1}{n}}_{\frac{1}{n+1}}\sin^{-1}(nx)dx.$ Then $\lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}}=$

Bati · 22. 12. 2018 12:07

Hi ↑ stuart clark:,
we can use a substitution to remove the n-dependence of the integrands and, with the fundamental theorem of calculus in mind, do some algebraic manipulation:

$\frac{a_n}{b_n}=\frac{\int_0^{\frac1n}\tan^{-1}(nx)-\int_0^{\frac1n(1-\frac1{n+1})}\tan^{-1}(nx)}{\int_0^{\frac1n}\sin^{-1}(nx)-\int_0^{\frac1n(1-\frac1{n+1})}\sin^{-1}(nx)}\\ =\frac{\frac1n\int^1_{1-\frac1{n+1}}\tan^{-1}y}{\frac1n\int^1_{1-\frac1{n+1}}\sin^{-1}y}\\ =\frac{(n+1)\int^1_{1-\frac1{n+1}}\tan^{-1}y}{(n+1)\int^1_{1-\frac1{n+1}}\sin^{-1}y}$ .
Since the integrands are left-continuous in 1, we can apply the fundamental theorem of calculus to obtain
$\lim_{n\to\infty}\frac{a_n}{b_n}=\frac{\tan^{-1}1}{\sin^{-1}1}=\frac12$ .

stuart clark · 24. 12. 2018 06:29

Thanks ↑ Bati:

Matematické Fórum

#1 22. 12. 2018 09:02

ratio of limits

#2 22. 12. 2018 12:07

Re: ratio of limits

#3 24. 12. 2018 06:29

Re: ratio of limits

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