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Question Number 194968 by cortano12 last updated on 21/Jul/23

Answered by tri26112004 last updated on 21/Jul/23

= lim_(x→0) (1/( (√(x+1))+(√(1−x)))).−lim_(x→0) (((√(x+1))+(√(1−x))−2)/x^2 ) =−(1/2).{lim_(x→0) (((1/2)[(√(4x+4))−(x+2)]+(√(1−x))+(1/2)x−1)/x^2 )} =−(1/2).[lim_(x→0) (1/2).((−1)/( (√(4x+4))+x+2))+lim_(x→0) (1/4).(1/( ((1/2)x−1)−(√(1−x))))] = −(1/2).((1/2).((−1)/4)+(1/4).(1/(−2))) = (1/8)

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

Answered by dimentri last updated on 21/Jul/23

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

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

Commented by mathlove last updated on 22/Jul/23

whay (((√(1+x))−(√(1−x))−x)/x^3 )=((((2x)/( (√(1+x))+(√(1−x))))−x)/x^3 )

$${whay}\: \\ $$$$\frac{\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}−{x}}{{x}^{\mathrm{3}} }=\frac{\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}}−{x}}{{x}^{\mathrm{3}} } \\ $$

Commented by dimentri last updated on 22/Jul/23

(((√(1+x))−(√(1−x)) .(((√(1+x)) +(√(1−x)))/( (√(1+x))+(√(1−x)))) −x)/x^3 ) = (((((1+x)−(1−x))/( (√(1+x)) +(√(1−x)))) −x)/x^3 )

$$\:\:\frac{\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}\:.\frac{\sqrt{\mathrm{1}+{x}}\:+\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}}\:−{x}}{{x}^{\mathrm{3}} } \\ $$$$\:=\:\frac{\frac{\left(\mathrm{1}+{x}\right)−\left(\mathrm{1}−{x}\right)}{\:\sqrt{\mathrm{1}+{x}}\:+\sqrt{\mathrm{1}−{x}}}\:−{x}}{{x}^{\mathrm{3}} } \\ $$

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