Question Number 196806 by cortano12 last updated on 01/Sep/23
$$\:\:\:\:\:\cancel{\underline{\underbrace{ }}} \\ $$
Answered by dimentri last updated on 01/Sep/23
$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{1}/\mathrm{6}} \left(\mathrm{2}−\mathrm{2}{x}\right)^{\mathrm{1}/\mathrm{6}} −{x}}{\mathrm{2}{x}\:−{x}\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{1}}+\:\sqrt[{\mathrm{3}}]{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }}}}\: \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}−\mathrm{2}{x}\right)^{\mathrm{1}/\mathrm{6}} −{x}^{\mathrm{5}/\mathrm{6}} }{{x}^{\mathrm{5}/\mathrm{6}} \:\left(\mathrm{2}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{1}+\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +}\:+\sqrt[{\mathrm{3}}]{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }}}\right.} \\ $$$$\:\:=\:\infty \\ $$
Commented by jabarsing last updated on 02/Sep/23
$${L}^{+} \left(\mathrm{0}\right):\:+\infty \\ $$$${L}^{−} \left(\mathrm{0}\right):\:{Not}\:{Exist} \\ $$
Answered by jabarsing last updated on 01/Sep/23
$${Not}\:{Exist} \\ $$$$ \\ $$$${because}\:{x}\rightarrow\mathrm{0}^{−} \:\:\:\Rightarrow\:\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\:\sqrt{\circleddash}\: \\ $$