Question Number 134173 by mathlove last updated on 28/Feb/21
$$\underset{{x}\rightarrow\mathrm{8}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{{x}}−\mathrm{2}}{\:\sqrt{{x}}−\mathrm{2}\sqrt{\mathrm{2}}}=? \\ $$
Answered by malwan last updated on 28/Feb/21
$$\underset{{x}\rightarrow\mathrm{8}} {{lim}}\:\frac{\:^{\mathrm{3}} \sqrt{{x}}−\mathrm{2}}{\:\sqrt{{x}}−\mathrm{2}\sqrt{\mathrm{2}}}×\frac{\:^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} }+\mathrm{2}\:^{\mathrm{3}} \sqrt{{x}}+\mathrm{4}}{\:^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} }+\mathrm{2}\:\:^{\mathrm{3}} \sqrt{{x}}+\mathrm{4}}×\frac{\sqrt{{x}}+\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{{x}}+\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$=\:\frac{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}+\mathrm{4}+\mathrm{4}}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{4}×\mathrm{3}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$
Answered by EDWIN88 last updated on 02/Mar/21
$$\mathrm{let}\:\mathrm{x}\:=\:\mathrm{t}^{\mathrm{6}} \:\Rightarrow\underset{\mathrm{t}\rightarrow\sqrt{\mathrm{2}}} {\mathrm{lim}}\:\:\:\left[\:\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{2}}{\mathrm{t}^{\mathrm{3}} −\mathrm{2}\sqrt{\mathrm{2}}}\:\right]\:=\:\underset{\mathrm{t}\rightarrow\sqrt{\mathrm{2}}} {\mathrm{lim}}\:\:\:\frac{\mathrm{2t}}{\mathrm{3t}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}}{\mathrm{3}×\sqrt{\mathrm{2}}} \\ $$$$\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$
Commented by malwan last updated on 02/Mar/21
$$\underset{{t}\rightarrow\sqrt{\mathrm{2}}} {{lim}}\:\:\frac{{t}^{\mathrm{2}} −\mathrm{2}}{{t}^{\mathrm{3}} −\mathrm{2}\sqrt{\mathrm{2}}}\:=\:\underset{{t}\rightarrow\sqrt{\mathrm{2}}} {{lim}}\:\frac{\left({t}−\sqrt{\mathrm{2}}\right)\left({t}+\sqrt{\mathrm{2}}\right)}{\left({t}−\sqrt{\mathrm{2}}\right)\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{2}\right)} \\ $$$$=\:\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}}{\mathrm{2}+\mathrm{2}+\mathrm{2}}\:=\:\frac{\mathrm{2}×\sqrt{\mathrm{2}}}{\mathrm{3}×\mathrm{2}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$