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Question Number 5427 by Rasheed Soomro last updated on 14/May/16 $$\mathrm{Sum}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}\:\mathrm{to}\:\mathrm{infinity}: \\ $$$$\mathrm{1}+\left(\mathrm{1}+\mathrm{k}\right)\mathrm{x}+\left(\mathrm{1}+\mathrm{k}+\mathrm{k}^{\mathrm{2}} \right)\mathrm{x}^{\mathrm{2}} +… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\:\:\:\mathrm{and}\:\:\:\:\mathrm{k}\:\:\:\mathrm{being}\:\mathrm{proper}\:\mathrm{fractions}. \\ $$ Commented by Yozzii last updated on…
Question Number 70913 by TawaTawa last updated on 09/Oct/19 Commented by mathmax by abdo last updated on 10/Oct/19 $${we}\:{have}\:\frac{{z}^{\mathrm{2017}} −\mathrm{1}}{{z}−\mathrm{1}}\:=\prod_{{k}=\mathrm{1}} ^{\mathrm{2016}} \left({z}−{e}^{\frac{{i}\mathrm{2}{k}\pi}{\mathrm{2017}}} \right) \\ $$$${z}=−\mathrm{1}\:\Rightarrow\mathrm{1}\:=\prod_{{k}=\mathrm{1}}…
Question Number 70886 by Omer Alattas last updated on 09/Oct/19 Commented by mathmax by abdo last updated on 10/Oct/19 $$\mathrm{1}−{cosu}\:\sim\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:{and}\:{sinu}\:\sim\:{u}\:{if}\:{u}\in{V}\left(\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{1}−{cos}\left(\frac{\pi{t}}{\mathrm{3}}\right)\sim\frac{\left(\frac{\pi{t}}{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{2}}\:=\pi^{\mathrm{2}} \frac{{t}^{\mathrm{2}}…
Question Number 5343 by sanusihammed last updated on 09/May/16 $$\left[\frac{\mathrm{1}−{x}^{−\mathrm{1}} −\mathrm{6}{x}^{−\mathrm{2}} }{\mathrm{1}−\mathrm{5}{x}^{−\mathrm{1}} −\mathrm{24}{x}^{−\mathrm{2}} }\right]^{{x}} \\ $$$$ \\ $$$${Limit}\:{x}\:\rightarrow\:\infty \\ $$$$ \\ $$$${please}\:{help}. \\ $$ Commented…
Question Number 70865 by OmerAlattas last updated on 08/Oct/19 Commented by mathmax by abdo last updated on 08/Oct/19 $$=\frac{\mathrm{0}}{\infty}=\mathrm{0} \\ $$ Commented by kaivan.ahmadi last…
Question Number 5286 by FilupSmith last updated on 05/May/16 $$\mathrm{Can}\:\mathrm{someone}\:\mathrm{please}\:\mathrm{explain}\:\mathrm{why}: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left({x}−{x}^{\mathrm{2}} \right)\:=\:−\infty \\ $$ Commented by FilupSmith last updated on 05/May/16 $$\mathrm{i}\:\mathrm{normally}\:\mathrm{undwrstand}\:\mathrm{limits}\:\mathrm{but} \\…
Question Number 5207 by FilupSmith last updated on 30/Apr/16 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{S}−\left(\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}!\right)}{{n}}−\mathrm{1}={L} \\ $$$${S}\in\mathbb{R},\:\:{S}>\mathrm{0} \\ $$$$ \\ $$$$\mathrm{What}\:\mathrm{can}\:\mathrm{we}\:\mathrm{tell}\:\mathrm{about}\:{L}? \\ $$$$ \\ $$$$\mathrm{Does}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{a}\:\mathrm{limit}? \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{positive}/\mathrm{negative}?…
Question Number 136278 by adhigenz last updated on 20/Mar/21 $$\mathrm{How}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}? \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\underset{\mathrm{1}} {\overset{{x}} {\int}}\left[\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}}\:−\:{x}\right]\:{dx}}{{x}^{\mathrm{2}} }\:=\:… \\ $$ Answered by MJS_new last updated on…