Question Number 123375 by bemath last updated on 25/Nov/20 $$\:\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\left(\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{sec}\:{x}\:\mathrm{tan}\:{x}\right)\:=? \\ $$ Answered by Dwaipayan Shikari last updated on 25/Nov/20 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\frac{\mathrm{1}−{sinx}}{{cos}^{\mathrm{2}} {x}}\right)=\frac{\mathrm{2}\left({sin}^{\mathrm{2}}…
Question Number 123349 by bemath last updated on 25/Nov/20 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\:{x}\:\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{9}{x}+\mathrm{2}}}\:−\frac{{x}}{\mathrm{3}}\:\right]? \\ $$ Answered by liberty last updated on 25/Nov/20 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\:\left[\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{9}{x}+\mathrm{2}}}\:−\:\frac{\mathrm{1}}{\mathrm{3}}\:\right]= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}\:\left(\mathrm{3}\sqrt{{x}−\mathrm{1}}\:−\sqrt{\mathrm{9}{x}+\mathrm{2}}\right)}{\:\mathrm{3}\sqrt{\mathrm{9}{x}+\mathrm{2}}}\:=…
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Question Number 188826 by mathlove last updated on 08/Mar/23 $${prove}\:{that} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt[{{n}\left({n}+\mathrm{1}\right)}]{\mathrm{1}!\:\mathrm{2}!\:\mathrm{3}!\centerdot\centerdot\centerdot\centerdot\centerdot{n}!}}{\:\sqrt{{n}}}={e}^{\frac{−\mathrm{3}}{\mathrm{4}}} \\ $$ Commented by Frix last updated on 07/Mar/23 $$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:{x}\:\mathrm{and}\:\mathrm{why}\:\rightarrow\mathrm{0}? \\ $$…
Question Number 188819 by mnjuly1970 last updated on 07/Mar/23 $$ \\ $$$$\:\:\:\:\:\:\:\mathrm{calculate}\: \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{lim}_{\:{x}\rightarrow\:\mathrm{0}^{\:+} } \left(\:\sqrt{\:\mathrm{cos}\:\left(\:\sqrt{{x}}\:\right)}\:\right)^{\:\mathrm{cot}\left(\:{x}\:\right)} \:=\:?\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$…
Question Number 188798 by mathlove last updated on 07/Mar/23 Answered by ARUNG_Brandon_MBU last updated on 07/Mar/23 $$\mathscr{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}^{{x}} \mathrm{3}^{{x}} \mathrm{4}^{{x}} \mathrm{5}^{{x}} …{n}^{{x}} −\mathrm{1}}{{x}} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}}…
Question Number 188771 by mnjuly1970 last updated on 06/Mar/23 Commented by mahdipoor last updated on 06/Mar/23 $${lim}\:\left({x}\rightarrow\mathrm{0}^{+} \right)\:\:\:\:\:\:\sqrt{{cos}\left(\sqrt{{cos}\left({x}\right)}\right)}=\sqrt{{cos}\left(\mathrm{1}\right)}={a} \\ $$$$\Rightarrow\mathrm{0}<{a}<\frac{\mathrm{1}}{\mathrm{2}}\overset{{b}>\mathrm{0}} {\Rightarrow}\mathrm{0}^{{b}} <{a}^{{b}} <\mathrm{1}^{{b}} \Rightarrow\mathrm{0}<{a}^{{b}} <\frac{\mathrm{1}}{\mathrm{2}^{{b}}…
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Question Number 188699 by cortano12 last updated on 05/Mar/23 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{sec}\:\mathrm{x}\:\mathrm{tan}\:\mathrm{x}\:\right)=? \\ $$ Answered by mehdee42 last updated on 05/Mar/23 $${li}\underset{{x}\rightarrow\infty} {{m}}\left(\frac{\mathrm{1}−{sinx}}{{cos}^{\mathrm{2}} {x}}\right)={li}\underset{{x}\rightarrow\infty} {{m}}\left(\frac{\mathrm{1}}{\mathrm{1}+{sinx}}\right)={A}…
Question Number 188575 by mathlove last updated on 03/Mar/23 Answered by universe last updated on 03/Mar/23 $$\sqrt{{e}} \\ $$ Answered by floor(10²Eta[1]) last updated on…