Question Number 9057 by sandipkd@ last updated on 16/Nov/16
Answered by aydnmustafa1976 last updated on 16/Nov/16
$${nsin}\frac{\mathrm{1}}{{n}}={lim}\frac{{sin}\frac{\mathrm{1}}{{n}}}{\frac{\mathrm{1}}{{n}}}={lim}\frac{{sint}}{{t}}=\mathrm{1}\:{therefore}\:\:\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\mathrm{4}.{arctgx}\mid_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{4}\left(\frac{\Pi}{\mathrm{4}}−\mathrm{0}\right)=\Pi \\ $$
Commented by sandipkd@ last updated on 17/Nov/16
$${thanks}..{i}\:{was}\:{confused}\:{in}\:{n}\mathrm{sin}\:\mathrm{1}/{n} \\ $$
Answered by aydnmustafa1976 last updated on 17/Nov/16
$${lim}\underset{{n}\rightarrow\infty} {\:}\int_{\mathrm{0}} ^{{nsin}\frac{\mathrm{1}}{{n}}} …..\:=\int_{\mathrm{0}} ^{{lim}\frac{\mathrm{1}}{\frac{\mathrm{1}}{{n}}}.{sin}\frac{\mathrm{1}}{{n}}} ….=\int_{\mathrm{0}} ^{{lim}\frac{{sin}\frac{\mathrm{1}}{{n}}}{\frac{\mathrm{1}}{{n}}}} …=…. \\ $$
Answered by aydnmustafa1976 last updated on 17/Nov/16
$${note}:\:{li}\underset{{n}\rightarrow\infty} {{m}}\frac{{sin}\frac{\mathrm{1}}{{n}}}{\frac{\mathrm{1}}{{n}}}={l}\underset{\frac{\mathrm{1}}{{n}}\rightarrow\mathrm{0}} {{i}m}\frac{{sin}\frac{\mathrm{1}}{{n}}}{\frac{\mathrm{1}}{{n}}}={li}\underset{{t}\rightarrow\mathrm{0}} {{m}}\frac{{sint}}{{t}}=\mathrm{0} \\ $$
Answered by aydnmustafa1976 last updated on 17/Nov/16
$${sorry}\:{lim}\frac{{sint}}{{t}}=\mathrm{1}\:{when}\:{t}\rightarrow\mathrm{0} \\ $$