Question Number 120277 by bemath last updated on 30/Oct/20
$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{3}} \:\left\{\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:−\:{x}\sqrt{\mathrm{2}}\:\right\}\:? \\ $$
Commented by benjo_mathlover last updated on 30/Oct/20
$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{4}} \left\{\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}\:−\sqrt{\mathrm{2}}\:\right\} \\ $$$$\:{setting}\:\frac{\mathrm{1}}{{x}}\:=\:{z}\:\wedge\:{z}\rightarrow\mathrm{0} \\ $$$$\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{z}^{\mathrm{4}} }}−\sqrt{\mathrm{2}}}{{z}^{\mathrm{4}} }\:×\:\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{z}^{\mathrm{4}} }}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{z}^{\mathrm{4}} }}+\sqrt{\mathrm{2}}} \\ $$$$\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+{z}^{\mathrm{4}} }−\mathrm{1}}{{z}^{\mathrm{4}} }\:×\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}×\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{z}^{\mathrm{4}} }{{z}^{\mathrm{4}} \left(\sqrt{\mathrm{1}+{z}^{\mathrm{4}} }+\mathrm{1}\right)} \\ $$$$=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:×\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{8}} \\ $$
Answered by Dwaipayan Shikari last updated on 30/Oct/20
$${x}^{\mathrm{3}} \left(\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}−{x}\sqrt{\mathrm{2}}\right) \\ $$$$={x}^{\mathrm{3}} \left(\sqrt{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} \sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}−{x}\sqrt{\mathrm{2}}\right) \\ $$$$={x}^{\mathrm{3}} \left({x}\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}−{x}\sqrt{\mathrm{2}}\right) \\ $$$$={x}^{\mathrm{4}} \left(\sqrt{\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{4}} }}−\sqrt{\mathrm{2}}\right) \\ $$$$=\sqrt{\mathrm{2}}{x}^{\mathrm{4}} \left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{4}} }}−\mathrm{1}\right) \\ $$$$=\sqrt{\mathrm{2}}\:{x}^{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{4}} }−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$
Answered by bemath last updated on 30/Oct/20
$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{3}} \:\left\{\frac{{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}−\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:+{x}\sqrt{\mathrm{2}}}\:\right\}\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} \:\left\{\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}−{x}^{\mathrm{2}} \:\right\}}{\:\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:+{x}\sqrt{\mathrm{2}}}\:=\: \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} \:\left\{{x}^{\mathrm{4}} +\mathrm{1}−{x}^{\mathrm{4}} \right\}}{{x}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}+\sqrt{\mathrm{2}}\:\right)\left(\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}+{x}^{\mathrm{2}} \:\right)}\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} \left(\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}+\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}+\mathrm{1}\right)}= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}+\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}+\mathrm{1}\right)}= \\ $$$$\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\right)×\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\: \\ $$$$\: \\ $$