Question Number 160793 by cortano last updated on 06/Dec/21 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}^{\mathrm{cos}\:\mathrm{x}} \:−\:\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\:=? \\ $$ Answered by qaz last updated on 06/Dec/21 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}^{\mathrm{cos}\:\mathrm{x}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}}…
Question Number 95230 by i jagooll last updated on 24/May/20 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{7}^{\sqrt{\mathrm{x}}} \:−\mathrm{1}}{\mathrm{2}^{\sqrt{\mathrm{x}}} \:−\mathrm{1}}\:=\:? \\ $$ Answered by bobhans last updated on 24/May/20 $$\underset{{x}\rightarrow\mathrm{0}^{+}…
Question Number 29689 by mrW2 last updated on 11/Feb/18 $${find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{2}}{\mathrm{1}+{a}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} =? \\ $$ Commented by abdo imad last updated on 13/Feb/18 $${let}\:{put}\:{A}\left({x}\right)=\left(\frac{\mathrm{2}}{\mathrm{1}+{a}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}}…
Question Number 160747 by amin96 last updated on 05/Dec/21 Commented by amin96 last updated on 05/Dec/21 $$ \\ $$does anyone have a solution? Answered by…
Question Number 95133 by john santu last updated on 23/May/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{x}^{\mathrm{2}} .\mathrm{e}^{\:−\mathrm{2x}} \right)\:=\:? \\ $$ Answered by bobhans last updated on 23/May/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}}…
Question Number 160609 by cortano last updated on 03/Dec/21 $$\:\:\:\:\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\left(\frac{\mathrm{tan}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\right)=? \\ $$ Answered by Ar Brandon last updated on 03/Dec/21 $$\mathscr{L}=\underset{{x}\rightarrow\pi} {\mathrm{lim}}\left(\frac{\mathrm{tan}{x}}{\mathrm{1}+\mathrm{cos}{x}}\right),\:{u}={x}−\pi \\ $$$$\:\:\:\:\:=\underset{{u}\rightarrow\mathrm{0}}…
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Question Number 160604 by cortano last updated on 03/Dec/21 $$\:\:\mathrm{S}_{\mathrm{n}} =\:\frac{\mathrm{12}}{\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)}+\frac{\mathrm{12}^{\mathrm{2}} }{\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{4}^{\mathrm{3}} −\mathrm{3}^{\mathrm{3}} \right)}+\frac{\mathrm{12}^{\mathrm{3}} }{\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{4}^{\mathrm{4}} −\mathrm{3}^{\mathrm{4}} \right)}+…+\frac{\mathrm{12}^{\mathrm{n}}…
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Question Number 29511 by abdo imad last updated on 09/Feb/18 $${find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:\:\:\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!}\:. \\ $$ Commented by prof Abdo imad last updated on 12/Feb/18 $${let}\:{use}\:{stirling}\:{formula}\:{we}\:{have}…
Question Number 29510 by abdo imad last updated on 09/Feb/18 $${let}\:\:\:{w}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left(\:\mathrm{1}\:+{e}^{\frac{\mathrm{1}}{{n}}} \:+{e}^{\frac{\mathrm{2}}{{n}}} \:+\:…\:{e}^{\frac{{n}−\mathrm{1}}{{n}}} \right)\:{find}\:{lim}_{{n}\rightarrow+\infty} {w}_{{n}} . \\ $$ Commented by abdo imad last updated…