Question Number 163386 by LEKOUMA last updated on 06/Jan/22
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\mathrm{1}−{x}^{\mathrm{3}} }−\sqrt[{\mathrm{4}}]{{x}^{\mathrm{4}} −\mathrm{1}} \\ $$
Answered by cortano1 last updated on 06/Jan/22
$$\:{Let}\:−{x}\:=\:\frac{\mathrm{1}}{{Y}}\:;\:\begin{cases}{{Y}\rightarrow\mathrm{0}}\\{{x}=−\frac{\mathrm{1}}{{Y}}}\end{cases} \\ $$$$\:\underset{{Y}\rightarrow\mathrm{0}} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\mathrm{1}−\left(−\frac{\mathrm{1}}{{Y}^{\mathrm{3}} }\right)}−\sqrt[{\mathrm{4}}]{\frac{\mathrm{1}}{{Y}^{\mathrm{4}} }−\mathrm{1}}\: \\ $$$$\:=\:\underset{{Y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{{Y}^{\mathrm{3}} +\mathrm{1}}−\sqrt[{\mathrm{4}}]{\mathrm{1}−{Y}^{\mathrm{4}} }}{{Y}} \\ $$$$\:=\:\underset{{Y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{{Y}^{\mathrm{3}} }{\mathrm{3}}\right)−\left(\mathrm{1}−\frac{{Y}^{\mathrm{4}} }{\mathrm{4}}\right)}{{Y}} \\ $$$$\:=\:\underset{{Y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{Y}^{\mathrm{2}} }{\mathrm{3}}\:+\:\frac{{Y}^{\mathrm{3}} }{\mathrm{4}}\:=\:\mathrm{0} \\ $$