Question Number 69906 by Tinku Tara last updated on 28/Sep/19 $$\mathrm{Notifications}\:\mathrm{are}\:\mathrm{working}\:\mathrm{now}. \\ $$$$\mathrm{We}\:\mathrm{are}\:\mathrm{still}\:\mathrm{working}\:\mathrm{on}\:\mathrm{other}\:\mathrm{issues}. \\ $$$$ \\ $$$$\mathrm{No}\:\mathrm{app}\:\mathrm{update}\:\mathrm{is}\:\mathrm{needed}.\:\mathrm{The}\:\mathrm{changes} \\ $$$$\mathrm{were}\:\mathrm{only}\:\mathrm{on}\:\mathrm{server}\:\mathrm{side}. \\ $$ Commented by ajfour last…
Question Number 135443 by 0731619177 last updated on 13/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 135442 by 0731619177 last updated on 13/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 135437 by 0731619177 last updated on 13/Mar/21 Answered by EDWIN88 last updated on 13/Mar/21 $$\mathrm{L}'\mathrm{H}\hat {\mathrm{o}pital}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{4x}−\mathrm{2sin}\:\mathrm{2x}}{\mathrm{2x}^{\mathrm{2}} +\mathrm{cos}\:\mathrm{2x}}}{\mathrm{4x}^{\mathrm{3}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} +\mathrm{cos}\:\mathrm{2x}}.\underset{{x}\rightarrow\mathrm{0}}…
Question Number 4365 by Filup last updated on 13/Jan/16 $$\mathrm{Is}\:\mathrm{the}\:\mathrm{following}\:\mathrm{correct}? \\ $$$$ \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}^{{i}−\mathrm{1}} \\ $$$${S}=\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{8}+\mathrm{16}+\mathrm{32}+\mathrm{64}+\mathrm{128}+… \\ $$$$\therefore\mathrm{2}{S}=\mathrm{2}+\mathrm{4}+\mathrm{8}+\mathrm{6}+\mathrm{32}+… \\ $$$$\mathrm{2}{S}={S}−\mathrm{1} \\ $$$${S}=−\mathrm{1} \\…
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Question Number 69901 by Rio Michael last updated on 28/Sep/19 Commented by Tony Lin last updated on 30/Sep/19 $$\left(\mathrm{1}\right){V}={I}\left({R}_{{A}} +{R}\right) \\ $$$$\because{so}\:{we}\:{may}\:{use}\:{the}\:{first}\:{circuit}\:{to} \\ $$$${measure}\:{something}\:{with}\:{high}\:{resistence} \\…
Question Number 135433 by naka3546 last updated on 13/Mar/21 $${Prove}\:\:{a}\:+\:\frac{\mathrm{1}}{{a}}\:\geqslant\:\mathrm{2}\:\:{for}\:\:{a}\:>\:\mathrm{0}\:\:{algebraically}\:. \\ $$ Answered by mr W last updated on 13/Mar/21 $${a}+\frac{\mathrm{1}}{{a}}=\left(\sqrt{{a}}−\frac{\mathrm{1}}{\:\sqrt{{a}}}\right)^{\mathrm{2}} +\mathrm{2}\geqslant\mathrm{2} \\ $$ Commented…
Question Number 4363 by Rasheed Soomro last updated on 13/Jan/16 $${If}\:\:\underset{{x}\rightarrow{a}} {{lim}}\:\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\:{exists}\:{and}\:\underset{{x}\rightarrow{a}} {{lim}}\:{g}\left({x}\right)=\mathrm{0}, \\ $$$${then}\:{show}\:{that} \\ $$$$\underset{{x}\rightarrow{a}} {{lim}}\:{f}\left({x}\right)=\mathrm{0}. \\ $$ Answered by Yozzii last updated…
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