Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 204477 by universe last updated on 18/Feb/24

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(2\mathrm{n}{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}}{1+{\mathrm{x}}^2}\mathrm{dx}\right)}^{\mathrm{n}}=?$$

Answered by witcher3 last updated on 18/Feb/24

$${\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}}{(1+{\mathrm{x}}^2)}\mathrm{dx}={\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}-{\mathrm{x}}^{\mathrm{n}+2}}{1-{\mathrm{x}}^4}\mathrm{dx}$$$$=\frac{1}{4}{\int }_0^1\frac{{\mathrm{t}}^{\frac{\mathrm{n}}{4}-\frac{3}{4}}-{\mathbf{t}}^{\frac{\mathbf{n}+2}{4}-\frac{3}{4}}}{1-\mathrm{t}}\mathbf{dt}=\frac{1}{4}(\mathbf{\Psi }(\frac{\mathrm{n}}{4}+\frac{3}{4})-\mathrm{\Psi }(\frac{\mathrm{n}}{4}+\frac{1}{4}))=\mathrm{f}\left(\mathrm{n}\right)$$$$\mathrm{\Psi }\left(\mathrm{z}\right)=\ln \left(\mathrm{z}\right)-\frac{1}{2\mathrm{z}}-\frac{1}{12\left({\mathrm{z}}^2\right)}+\mathrm{o}\left(\frac{1}{{\mathrm{z}}^3}\right)$$$$\mathrm{\Psi }\left(\frac{\mathrm{n}+3}{4}\right)-\mathrm{\Psi }\left(\frac{\mathrm{n}+1}{4}\right)=\ln \left(\frac{\mathrm{n}+3}{\mathrm{n}+1}\right)-\frac{1}{2}(\frac{4}{\mathrm{n}+3}-\frac{4}{\mathrm{n}+1})-\frac{1}{12}(\frac{16}{{(\mathrm{n}+3)}^2}-\frac{16}{{(\mathrm{n}+1)}^2})+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^3}\right)$$$$\ln (1+\frac{3}{\mathrm{n}})-\ln (1+\frac{1}{\mathrm{n}})-\frac{2}{\mathrm{n}}(\frac{1}{1+\frac{3}{\mathrm{n}}}-\frac{1}{1+\frac{1}{\mathrm{n}}})-\frac{4}{{3\mathrm{n}}^2}(\frac{1}{{(1+\frac{3}{\mathrm{n}})}^2}-\frac{1}{{(1+\frac{1}{\mathrm{n}})}^2})$$$$\frac{1}{{(1+\mathrm{x})}^2}=\sum _{\mathrm{k}⩾1}{(-1)}^{\mathrm{k}-1}{\mathrm{kx}}^{\mathrm{k}-1}$$$$=\frac{2}{\mathrm{n}}-\frac{4}{{\mathrm{n}}^2}-\frac{2}{\mathrm{n}}(1-\frac{3}{\mathrm{n}}+\frac{9}{{\mathrm{n}}^2}-(1-\frac{1}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^2}))-\frac{4}{{3\mathrm{n}}^2}(1-\frac{6}{\mathrm{n}}-(1-\frac{2}{\mathrm{n}}))+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^3}\right)$$$$=\frac{2}{\mathrm{n}}-\frac{16}{{\mathrm{n}}^3}+\frac{16}{{3\mathrm{n}}^3}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^3}\right)$$$$=\frac{2}{\mathrm{n}}-\frac{32}{{3\mathrm{n}}^3}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^3}\right)$$$$\frac{\mathrm{n}}{2}(\frac{2}{\mathrm{n}}-\frac{32}{{3\mathrm{n}}^3}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^3}\right))=1-\frac{16}{{3\mathrm{n}}^2}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right)$$$${\mathrm{e}}^{\mathrm{nln}(1-\frac{16}{{3\mathrm{n}}^2}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right))}={\mathrm{e}}^{-\frac{16}{3\mathrm{n}}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)}\to 1$$

Commented by universe last updated on 19/Feb/24

$$thankyousomuchsir$$

Commented by witcher3 last updated on 19/Feb/24

$$\mathrm{withe}\mathrm{pleasur}\mathrm{i}\mathrm{think}\mathrm{we}\mathrm{can}\mathrm{prove}\mathrm{it}\mathrm{withe}\mathrm{high}\mathrm{school}\mathrm{tools}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com