Question Number 120312 by mnjuly1970 last updated on 30/Oct/20 $$\:\:\:\:\:\:\:\:\:\:…\:\clubsuit{nice}\:\:{calculus}\clubsuit… \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:{lim}_{{n}\rightarrow\infty} \frac{\sqrt[{{n}^{\mathrm{2}} }]{{n\$}}}{\:\sqrt{{n}}}?\overset{???} {=}{e}^{\frac{−\mathrm{3}}{\mathrm{4}}} \\ $$$$\:\:\:\:\:{where}\:::\:\:{n\$}\:\overset{{superfactorial}} {=}{n}!.\left({n}−\mathrm{1}\right)!.\left({n}−\mathrm{2}\right)!…\mathrm{3}!.\mathrm{2}!.\mathrm{1}! \\ $$$$\:\:\:\:\:\:\:\:…\spadesuit{m}.{n}.\mathrm{1970}\spadesuit… \\…
Question Number 120243 by benjo_mathlover last updated on 30/Oct/20 Answered by bemath last updated on 30/Oct/20 $${let}\:{f}\left({x}\right)\:=\:{x}^{\frac{\mathrm{1}}{{x}^{\mathrm{6}} }} \:\Leftrightarrow\:\mathrm{ln}\:{f}\left({x}\right)=\:\frac{\mathrm{ln}\:\left({x}\right)}{{x}^{\mathrm{6}} } \\ $$$${differentiating}\:{both}\:{sides}\:{gives} \\ $$$$\frac{{f}\:'\left({x}\right)}{{f}\left({x}\right)}\:=\:\frac{{x}^{\mathrm{5}} −\mathrm{6}{x}^{\mathrm{5}}…
Question Number 120233 by MehraGanesh last updated on 30/Oct/20 $$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 54679 by Raj Singh last updated on 09/Feb/19 Commented by maxmathsup by imad last updated on 09/Feb/19 $${let}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{{x}+{a}}}\:\:{for}\:{x}>−{a}\:\:{we}\:{have}\:{f}^{'} \left({x}\right)={lim}_{{h}\rightarrow\mathrm{0}} \:\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \:\:\frac{\frac{\mathrm{1}}{\:\sqrt{{x}+{h}+{a}}\:\:}−\frac{\mathrm{1}}{\:\sqrt{{x}+{a}}}}{{h}}\:={lim}_{{h}\rightarrow\mathrm{0}}…
Question Number 119969 by huotpat last updated on 28/Oct/20 Answered by Dwaipayan Shikari last updated on 28/Oct/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{k}^{\mathrm{2}} +{k}+\mathrm{1}} \\ $$$$\underset{{k}=\mathrm{1}}…
Question Number 119965 by sdfg last updated on 28/Oct/20 Answered by Dwaipayan Shikari last updated on 28/Oct/20 $$\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$\frac{{d}\Gamma\left({x}\right)}{{dx}}=\int_{\mathrm{0}} ^{\infty}…
Question Number 119937 by bobhans last updated on 28/Oct/20 $$\:\left({i}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{cos}\:\frac{\mathrm{9}\pi}{\mathrm{7}}\right)? \\ $$$$\left({ii}\right)\:\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{1}°\right)\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{2}°\right)\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{3}°\right)×…×\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{29}°\right)? \\ $$ Answered by TANMAY PANACEA last updated on 28/Oct/20 $$\left.{ii}\right)\:\sqrt{\mathrm{3}}\:+{tan}\mathrm{1}^{{o}} \\ $$$$={tan}\mathrm{60}^{{o}}…
Question Number 185453 by cortano1 last updated on 22/Jan/23 $$\:{Find}\:{the}\:{largest}\:{possible}\:{area} \\ $$$$\:{of}\:{trapezoid}\:{that}\:{can}\:{be}\:{drawn}\: \\ $$$$\:{under}\:{the}\:{x}−{axis}\:{so}\:{that}\:{one}\: \\ $$$$\:{of}\:{its}\:{bases}\:{is}\:{on}\:{the}\:{x}−{axis}\: \\ $$$$\:{and}\:{the}\:{other}\:{two}\:{vertices}\:{are} \\ $$$$\:{on}\:{the}\:{curve}\:{y}={x}^{\mathrm{2}} −\mathrm{9}\: \\ $$ Commented by…
Question Number 119724 by benjo_mathlover last updated on 26/Oct/20 $${Let}\:{x},{y},{z}\:{be}\:{non}\:{negative}\:{real}\:{numbers} \\ $$$${such}\:{that}\:{x}+{y}+{z}=\mathrm{1}.\:{Find}\:{the}\:{extremum} \\ $$$${of}\:{F}\:=\:\mathrm{2}{x}^{\mathrm{2}} +{y}+\mathrm{3}{z}^{\mathrm{2}} \:. \\ $$ Answered by mindispower last updated on 26/Oct/20…
Question Number 119692 by bemath last updated on 26/Oct/20 $${If}\:{x}\:{is}\:{real}\:{number}\:{satisfying} \\ $$$$\mathrm{3}{x}+\frac{\mathrm{1}}{\mathrm{2}{x}}=\mathrm{4}\:,\:{find}\:{the}\:{value}\:{of} \\ $$$$\mathrm{27}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{3}} }\:. \\ $$ Commented by bemath last updated on 26/Oct/20…