Question Number 133068 by mnjuly1970 last updated on 18/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{advanced}…..{calculus}…. \\ $$$$\:\:\:{evaluation}::\:\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \left(\:\frac{\zeta\left({k}\right)−\mathrm{1}}{{k}}\right) \\ $$$$\:\:\:\::::\boldsymbol{\Phi}=\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \:\frac{\zeta\left({k}\right)−\mathrm{1}}{{k}} \\ $$$$\:\:\:\:\:\:\:\:\:=\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:\underset{{n}=\mathrm{2}}…
Question Number 132971 by metamorfose last updated on 17/Feb/21 $$\:\:{Lim}_{{x}\rightarrow\left(\frac{\pi}{\mathrm{2}}\right)^{−\:\:\:} } \frac{\lfloor{sin}\left({x}\right)\rfloor}{{cos}\left({x}\lfloor{x}\rfloor\right)} \\ $$ Answered by mnjuly1970 last updated on 18/Feb/21 $${ans}:\frac{\mathrm{0}}{\mathrm{0}^{+} }=\mathrm{0} \\ $$…
Question Number 132928 by metamorfose last updated on 17/Feb/21 $$\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right) \\ $$ Commented by Olaf last updated on 17/Feb/21 $${sorry}\:{sir},\:{I}\:{deleted}\:{my}\:{answer}. \\ $$$${it}\:{was}\:{wrong}.…
Question Number 1817 by 112358 last updated on 05/Oct/15 $${Evaluate}\:{the}\:{following}\:{limit}, \\ $$$${if}\:{it}\:{exists}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\left({sin}\left({sinx}\right)\right)}{{x}}. \\ $$ Answered by 123456 last updated on 05/Oct/15 $${f}_{{n}}…
Question Number 1766 by Rasheed Ahmad last updated on 18/Sep/15 $${Determine}\: \\ $$$$\:\:\left({i}\right)\:\:\underset{{a}\rightarrow\infty} {{lim}}\:{a}^{\frac{\mathrm{1}}{{a}}} \:\:\:\:\:\:\:\:\:\:\:\left({ii}\right)\:\underset{{a}\rightarrow\mathrm{0}} {{lim}}\:{a}^{\frac{\mathrm{1}}{{a}}} \\ $$ Answered by 123456 last updated on 19/Sep/15…
Question Number 1767 by Rasheed Ahmad last updated on 18/Sep/15 $${Determine} \\ $$$$\:\left({i}\right)\:\underset{{a}\rightarrow\infty} {{lim}}\:\left(\frac{\mathrm{1}}{{a}}\right)^{{a}} \:\:\:\left({ii}\right)\:\:\underset{{a}\rightarrow\mathrm{0}} {{lim}}\:\left(\frac{\mathrm{1}}{{a}}\right)^{{a}} \: \\ $$ Answered by 123456 last updated on…
Question Number 1608 by 112358 last updated on 26/Aug/15 $${Find}\:{the}\:{limit}\:{of}\:{this}\:{sequence}. \\ $$$$\sqrt{\mathrm{2}},\sqrt{\mathrm{2}\sqrt{\mathrm{2}}},\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}},\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}}},… \\ $$$${Show}\:{that}\:{the}\:{sum}\:{of}\:{the} \\ $$$${terms}\:{of}\:{this}\:{infinite}\:{sequence} \\ $$$${does}\:{not}\:{converge}. \\ $$ Commented by Rasheed Ahmad last…
Question Number 1499 by 112358 last updated on 14/Aug/15 $${Use}\:{the}\:\epsilon−\delta\:{definition}\:{of}\:{the} \\ $$$${limit}\:{to}\:{show}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{{x}}{\mathrm{6}−{x}}=\mathrm{1}\:. \\ $$$$\left({I}'{m}\:{hoping}\:{to}\:{better}\:{understand}\right. \\ $$$${this}\:{concept}\:{by}\:{example}\:{so}\:{please} \\ $$$${help}\:{me}\:{by}\:{explaning}\:{the}\: \\ $$$$\left.{reasoning}\:{behind}\:{your}\:{steps}.\right) \\ $$…
Question Number 132550 by liberty last updated on 15/Feb/21 $$\mathrm{If}\:{a}=\mathrm{1}\:\mathrm{then}\:\underset{{x}\rightarrow\left({a}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{5}\right)} {\mathrm{lim}}\frac{−\mathrm{4}{x}^{\mathrm{2}} +\sqrt{\mathrm{3}{x}+\mathrm{1}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{8}{x}}+\mathrm{2}}\:=? \\ $$$$\left(\mathrm{a}\right)\:−\mathrm{1}\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:−\frac{\mathrm{4}}{\mathrm{5}}\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:−\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\left(\mathrm{d}\right)\:−\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\:\left(\mathrm{e}\right)\:−\frac{\mathrm{1}}{\mathrm{5}} \\ $$ Commented by MJS_new last updated…
Question Number 132483 by abdullahquwatan last updated on 14/Feb/21 $$\underset{{x}\rightarrow−\mathrm{2}} {\mathrm{lim}}\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}−{x}^{\mathrm{2}} +\mathrm{2}}{{x}^{\mathrm{5}} +\mathrm{32}}\:\mathrm{no}\:\mathrm{hospital} \\ $$ Commented by EDWIN88 last updated on 14/Feb/21 $$\mathrm{what}\:\mathrm{hospital}? \\…