Question Number 136670 by rexford last updated on 24/Mar/21 Answered by Dwaipayan Shikari last updated on 24/Mar/21 $$\mathrm{0} \\ $$ Answered by MJS_new last updated…
Question Number 5593 by sanusihammed last updated on 21/May/16 $${Find}\:{the}\:{limit} \\ $$$$ \\ $$$${limit}\:\:\:\:\:\:\:\left[\mathrm{1}\:+\:{mt}\right]^{\frac{\mathrm{2}}{{t}}} \\ $$$${t}\:\rightarrow\infty \\ $$ Answered by FilupSmith last updated on 21/May/16…
Question Number 71096 by Omer Alattas last updated on 11/Oct/19 Answered by $@ty@m123 last updated on 11/Oct/19 $${Let}\:{x}=\mathrm{1}+{y} \\ $$$$\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\:\frac{{y}}{\mathrm{ln}\:\left(\mathrm{1}+{y}\right)} \\ $$$$\:\:\frac{\mathrm{1}}{\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+{y}\right)}{{y}}}=\mathrm{1} \\…
Question Number 136594 by slahadjb last updated on 23/Mar/21 $${li}\underset{\mathrm{0}^{+} } {{m}}\frac{{x}^{{x}^{{x}} } }{{x}^{{x}} −\mathrm{1}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 5427 by Rasheed Soomro last updated on 14/May/16 $$\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…