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Question-196806




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

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