Question Number 163041 by qaz last updated on 03/Jan/22 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\sqrt{\mathrm{n}}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} }{\:\sqrt{\mathrm{n}}\mathrm{x}^{\mathrm{2}} }\mathrm{dx}−\sqrt{\pi}\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163040 by qaz last updated on 03/Jan/22 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{n}!}{\mathrm{n}^{\mathrm{n}} }\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{n}^{\mathrm{k}} }{\mathrm{k}!}−\underset{\mathrm{k}=\mathrm{n}+\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{n}^{\mathrm{k}} }{\mathrm{k}!}\right)=? \\ $$ Terms of Service Privacy Policy…
Question Number 163042 by qaz last updated on 03/Jan/22 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\sqrt{\mathrm{n}}\int_{−\infty} ^{+\infty} \frac{\mathrm{cos}\:\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} }\mathrm{dx}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163043 by qaz last updated on 03/Jan/22 $$\underset{\mathrm{x}\rightarrow−\infty} {\mathrm{lim}}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}^{\mathrm{n}} }=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163039 by qaz last updated on 03/Jan/22 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\sqrt{\mathrm{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}\centerdot\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{2}\pi\mathrm{x}\right)\mathrm{dx}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 97398 by RAMANA last updated on 07/Jun/20 $$\mathrm{The}\:\mathrm{discontinuty}\:\mathrm{of}\:\left[\mathrm{x}\right]^{\mathrm{2}} −\left[\mathrm{x}^{\mathrm{2}} \right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 162925 by qaz last updated on 02/Jan/22 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\left[\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}}} −\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}\right)^{\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}}} \right]=? \\…
Question Number 162926 by qaz last updated on 02/Jan/22 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }\left[\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}}{\mathrm{2}}+…+\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}}\right)^{\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{x}}{\mathrm{2}}+…+\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}}}} −\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}}{\mathrm{2}}+…+\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{x}}{\mathrm{2}}+…+\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}}} \right]=? \\ $$ Terms of Service Privacy Policy…
Question Number 97355 by bagjamath last updated on 07/Jun/20 Commented by PRITHWISH SEN 2 last updated on 07/Jun/20 $$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\:\left(\frac{\pi}{\mathrm{n}+\mathrm{1}}+\frac{\pi}{\mathrm{n}+\mathrm{1}}+……+\frac{\pi}{\mathrm{n}+\mathrm{1}}\:\mathrm{ntimes}\:\right) \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\pi}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}\:=\:\pi \\ $$…
Question Number 162827 by saboorhalimi last updated on 01/Jan/22 Answered by Ar Brandon last updated on 01/Jan/22 $$\mathrm{g}\left({x}\right)=\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\left({x}+\mathrm{1}\right)^{{r}+\mathrm{1}} −{x}^{{r}+\mathrm{1}} \right)^{\frac{\mathrm{1}}{{r}}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{g}\left({x}\right)}{{x}}=\underset{{r}\rightarrow\mathrm{0},\:{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\mathrm{1}+\frac{\mathrm{1}}{{r}}−\mathrm{1}}…