Question Number 187164 by mathlove last updated on 14/Feb/23 | ||
Answered by qaz last updated on 14/Feb/23 | ||
$$\underset{{p}\rightarrow{ln}\frac{\mathrm{1}}{\mathrm{2}}} {{lim}}\frac{{ln}\mathrm{2}+{ln}\mathrm{2}\:{cos}\:{p}}{\left({cos}\:{ln}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }=\frac{\left(\mathrm{1}+{cos}\:{ln}\mathrm{2}\right){ln}\mathrm{2}}{{cos}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}}=\frac{\left[\mathrm{2cos}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}\right]{ln}\mathrm{2}}{{cos}^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}}=\mathrm{2}{ln}\mathrm{2} \\ $$$$\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{p}} }{{p}+\mathrm{1}}=\underset{{p}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} }{{p}}={ln}\mathrm{2} \\ $$$$\frac{{d}}{{dx}}\left(\frac{{x}}{\left(\mathrm{sin}\:{x}\right)^{\mathrm{2}} +\left(\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\right)=\frac{{dx}}{{dx}}=\mathrm{1} \\ $$$$\underset{{t}\rightarrow\infty} {{lim}}\left(\mathrm{2}{t}+\mathrm{1}\right)\frac{\left({t}+\mathrm{2}\right)!+\left({t}+\mathrm{1}\right)!}{{t}+\mathrm{3}}\left(\frac{\mathrm{3}^{{t}} −\mathrm{2}^{{t}} }{\mathrm{2}^{{t}} +\mathrm{5}}\right) \\ $$$$=\underset{{t}\rightarrow\infty} {{lim}}\left(\mathrm{2}{t}+\mathrm{1}\right)\left({t}+\mathrm{1}\right)!\centerdot\frac{\mathrm{1}−\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{t}} }{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{t}} +\frac{\mathrm{5}}{\mathrm{3}^{{t}} }}=+\infty \\ $$$${I}=\int_{{ln}\mathrm{2}} ^{\mathrm{2}{ln}\mathrm{2}} \left(+\infty\right){dx}=? \\ $$ | ||