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Question Number 171033 by cortano1 last updated on 06/Jun/22

  lim_(x→0)  (1/x) [ ((((1−(√(1−x)))/( (√(1+x))−1)) ))^(1/3) −1 ]=?

$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}}\:\left[\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}−\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}−\mathrm{1}}\:}−\mathrm{1}\:\right]=? \\ $$

Commented by greougoury555 last updated on 07/Jun/22

 (((((1−(√(1−x)))(1+(√(1−x))))/(1+(√(1−x)))).(((√(1+x))+1)/(((√(1+x))−1)((√(1+x))+1)))))^(1/3)   = (((x((√(1+x))+1))/(x(1+(√(1−x))))))^(1/3)  = (((((√(1+x))+1)/(1+(√(1−x)))) ))^(1/3)   L=lim_(x→0)  (1/x) ((((((√(1+x))+1))^(1/3)  −((1+(√(1−x))))^(1/3) )/( ((1+(√(1−x))))^(1/3) )))   = (1/( (2)^(1/3) )) .lim_(x→0)  (((((√(1+x))+1))^(1/3) −((1+(√(1−x))))^(1/3) )/x)   =(1/( (2)^(1/3) )) .lim_(x→0)  ((((√(1+x))+1)−(1+(√(1−x))))/(x[ ((((√(1+x))+1)^2 ))^(1/3) +((((√(1+x))+1)(1+(√(1−x)))))^(1/3) +(((1+(√(1−x)))^2 ))^(1/3) ))    =(1/( 3.(8)^(1/3) )) .lim_(x→0)  (((√(1+x))−(√(1−x)))/x)   = (1/6).lim_(x→0)  ((2x)/(x((√(1+x))+(√(1−x)))))   = (1/6).(2/2)= (1/6)

$$\:\sqrt[{\mathrm{3}}]{\frac{\left(\mathrm{1}−\sqrt{\mathrm{1}−{x}}\right)\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}}\right)}{\mathrm{1}+\sqrt{\mathrm{1}−{x}}}.\frac{\sqrt{\mathrm{1}+{x}}+\mathrm{1}}{\left(\sqrt{\mathrm{1}+{x}}−\mathrm{1}\right)\left(\sqrt{\mathrm{1}+{x}}+\mathrm{1}\right)}} \\ $$$$=\:\sqrt[{\mathrm{3}}]{\frac{{x}\left(\sqrt{\mathrm{1}+{x}}+\mathrm{1}\right)}{{x}\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}}\right)}}\:=\:\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{1}+{x}}+\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{1}−{x}}}\:} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}}\:\left(\frac{\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+{x}}+\mathrm{1}}\:−\sqrt[{\mathrm{3}}]{\mathrm{1}+\sqrt{\mathrm{1}−{x}}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\sqrt{\mathrm{1}−{x}}}}\right) \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+{x}}+\mathrm{1}}−\sqrt[{\mathrm{3}}]{\mathrm{1}+\sqrt{\mathrm{1}−{x}}}}{{x}} \\ $$$$\:=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{1}+{x}}+\mathrm{1}\right)−\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}}\right)}{{x}\left[\:\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{1}+{x}}+\mathrm{1}\right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{1}+{x}}+\mathrm{1}\right)\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}}\right)}+\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}}\right)^{\mathrm{2}} }\right.}\: \\ $$$$\:=\frac{\mathrm{1}}{\:\mathrm{3}.\sqrt[{\mathrm{3}}]{\mathrm{8}}}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}}{{x}} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{6}}.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}}{{x}\left(\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}\right)} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{6}}.\frac{\mathrm{2}}{\mathrm{2}}=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Answered by qaz last updated on 06/Jun/22

lim_(x→0) (1/x)[(((1−(√(1−x)))/( (√(1+x))−1)))^(1/3) −1]  =lim_(x→0) (1/x)ln(((1−(√(1−x)))/( (√(1+x))−1)))^(1/3)   =lim_(x→0) (1/(3x))ln((1−(√(1−x)))/( (√(1+x))−1))  =lim_(x→0) (1/(3x))∙(((1−(√(1−x)))/( (√(1+x))−1))−1)  =lim_(x→0) ((2−(√(1−x))−(√(1+x)))/(3x((√(1+x))−1)))  =lim_(x→0) (((1/4)x^2 +o(x^2 ))/((3/2)x^2 +o(x^2 )))  =(1/6)

$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}}\left[\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{x}}}{\:\sqrt{\mathrm{1}+\mathrm{x}}−\mathrm{1}}}−\mathrm{1}\right] \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{x}}}{\:\sqrt{\mathrm{1}+\mathrm{x}}−\mathrm{1}}} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{3x}}\mathrm{ln}\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{x}}}{\:\sqrt{\mathrm{1}+\mathrm{x}}−\mathrm{1}} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{3x}}\centerdot\left(\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{x}}}{\:\sqrt{\mathrm{1}+\mathrm{x}}−\mathrm{1}}−\mathrm{1}\right) \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}−\sqrt{\mathrm{1}−\mathrm{x}}−\sqrt{\mathrm{1}+\mathrm{x}}}{\mathrm{3x}\left(\sqrt{\mathrm{1}+\mathrm{x}}−\mathrm{1}\right)} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{2}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)}{\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Answered by qaz last updated on 06/Jun/22

lim_(x→0) (1/x)[(((1−(√(1−x)))/( (√(1+x))−1)))^(1/3) −1]  =lim_(x→0) (1/x)[((((1/2)x+(1/8)x^2 +o(x^2 ))/((1/2)x−(1/8)x^2 +o(x^2 ))))^(1/3) −1]  =lim_(x→0) (1/x)[(((1+(1/4)x+o(x))/(1−(1/4)x+o(x))))^(1/3) −1]  =lim_(x→0) (1/x)[(((1+(1/4)x+o(x))(1+(1/4)x+o(x))))^(1/3) −1]  =lim_(x→0) (1/x)[((1+(1/2)x+o(x)))^(1/3) −1]  =lim_(x→0) (1/x)((1/6)x+o(x))  =(1/6)

$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}}\left[\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{x}}}{\:\sqrt{\mathrm{1}+\mathrm{x}}−\mathrm{1}}}−\mathrm{1}\right] \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}}\left[\sqrt[{\mathrm{3}}]{\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{x}^{\mathrm{2}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{x}^{\mathrm{2}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)}}−\mathrm{1}\right] \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}}\left[\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)}}−\mathrm{1}\right] \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}}\left[\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)\right)}−\mathrm{1}\right] \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}}\left[\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)}−\mathrm{1}\right] \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}}\left(\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

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