Question Number 83919 by john santu last updated on 08/Mar/20 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{3}}{\pi}\mathrm{arc}\:\mathrm{tan}\:{x}\right)^{\mathrm{2}{x}} \:=\:? \\ $$ Commented by john santu last updated on 08/Mar/20 $$=\:\mathrm{e}\:^{\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\left(\mathrm{ln}\left(\frac{\mathrm{3}}{\pi}\mathrm{tan}^{−\mathrm{1}}…
Question Number 83865 by jagoll last updated on 07/Mar/20 $$\underset{{x}\rightarrow−\infty\:} {\mathrm{lim}}\:\left(\mathrm{x}\sqrt{\mathrm{2x}+\mathrm{2}}−\mathrm{x}\sqrt{\mathrm{2x}+\mathrm{3}}\right) \\ $$ Commented by mr W last updated on 07/Mar/20 $$\mathrm{2}{x}+\mathrm{2}\geqslant\mathrm{0} \\ $$$$\Rightarrow{x}\geqslant−\mathrm{1} \\…
Question Number 83859 by john santu last updated on 06/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\:\mathrm{cot}\:^{\mathrm{2}} {x}\right)=\:? \\ $$ Commented by john santu last updated on 06/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 149377 by fotosy2k last updated on 05/Aug/21 Answered by john_santu last updated on 05/Aug/21 $$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}.\mathrm{3}^{\mathrm{n}} }{\mathrm{3}^{\mathrm{n}} \left(\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{n}} \right)}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}}{\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{n}} }=\mathrm{3} \\ $$…
Question Number 149372 by fotosy2k last updated on 05/Aug/21 Answered by john_santu last updated on 05/Aug/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\mathrm{x}+\mathrm{2sin}\:\mathrm{x}\right)}{\mathrm{sin}\:\mathrm{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{sin}\:\left(\mathrm{x}+\mathrm{2sin}\:\mathrm{x}\right)}{\mathrm{x}+\mathrm{2sin}\:\mathrm{x}}\:.\:\mathrm{x}+\mathrm{2sin}\:\mathrm{x}}{\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}.\:\mathrm{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}+\mathrm{2sin}\:\mathrm{x}}{\mathrm{x}}\:=\:\mathrm{1}+\mathrm{2}=\mathrm{3} \\…
Question Number 149373 by fotosy2k last updated on 05/Aug/21 Answered by john_santu last updated on 05/Aug/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{\mathrm{3x}}{\mathrm{2}}\right)−\left(\mathrm{1}+\frac{\mathrm{2x}}{\mathrm{2}}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{2x}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{x}}{\mathrm{2}}}{\mathrm{x}}\:×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2x}}=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Commented…
Question Number 149346 by Study last updated on 04/Aug/21 $${li}\underset{{x}\rightarrow\infty} {{m}}\left(\frac{\mathrm{3}{x}+\mathrm{12}}{\mathrm{6}{x}−\mathrm{20}}\right)^{\mathrm{2}{x}} =? \\ $$ Commented by Study last updated on 04/Aug/21 $${helpe}\:{me} \\ $$ Answered…
Question Number 83710 by john santu last updated on 05/Mar/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{{x}} −\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)^{{x}} }{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{{x}−\mathrm{1}} −\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)^{{x}−\mathrm{1}} }\:=\:? \\ $$ Commented by john santu last updated on…
Question Number 149251 by john_santu last updated on 04/Aug/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{tan}\:\mathrm{x}\:\sqrt{\mathrm{tan}\:\mathrm{x}}−\mathrm{sin}\:\mathrm{x}\:\sqrt{\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\mathrm{3}} \:\sqrt{\mathrm{x}}}\:=? \\ $$$$ \\ $$ Answered by dumitrel last updated on 04/Aug/21 $$\underset{{x}\rightarrow\mathrm{0}^{+}…
Question Number 149239 by john_santu last updated on 04/Aug/21 $$\:\:\:\underset{{x}\rightarrow\mathrm{a}^{+} } {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}}−\sqrt{\mathrm{a}}\:−\sqrt{\mathrm{x}−\mathrm{a}}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:=?\: \\ $$ Answered by EDWIN88 last updated on 04/Aug/21 $$\:\underset{{x}\rightarrow{a}^{+} }…