Question Number 220843 by Nicholas666 last updated on 20/May/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha\:\in\:\mathbb{R} \\ $$$$\:\:\:\:\:\mathrm{lim}_{{x}\rightarrow\mathrm{1}} \:\frac{\left(\mathrm{1}\:−\:{x}\right)^{\alpha} }{\:^{\mathrm{3}} \sqrt{\mathrm{1}\:−\:{x}^{\mathrm{4}} }}\:\:\:\:\:\:\:\:\in\left(\mathrm{0},\infty\right) \\ $$$$ \\ $$
Commented by SdC355 last updated on 20/May/25
$$ \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\left(\mathrm{1}−{x}\right)^{\alpha} }{\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\:^{\mathrm{3}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\alpha\left(\mathrm{1}−{x}\right)^{\alpha−\mathrm{1}} }{\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}\centerdot^{\mathrm{3}} \sqrt{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }}} \\ $$$$\therefore\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\frac{\mathrm{3}\alpha\left(\mathrm{1}−{x}\right)^{\alpha−\mathrm{1}} \centerdot^{\mathrm{3}} \sqrt{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }}{\mathrm{4}{x}^{\mathrm{3}} }=\mathrm{0} \\ $$