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Question Number 194812 by sonukgindia last updated on 16/Jul/23

Answered by MM42 last updated on 16/Jul/23

A=lim_(x→0) ((((√(1−x^2 ))/(1−x))−(1+x))/( ((sinx−x)/x))) A=lim_(x→0) ((x(√(1−x^2 ))(1−(√(1−x^2 ))))/( (1−x)(sinx−x))) =lim_(x→0) ((x(√(1−x^2 ))×x^2 )/( (1−x)×(1+(√(1−x^2 )))×−(1/6)x^3 )) =−(3/2)✓

$${A}={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{1}−{x}}−\left(\mathrm{1}+{x}\right)}{\:\frac{{sinx}−{x}}{{x}}} \\ $$$${A}={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\left(\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{\:\left(\mathrm{1}−{x}\right)\left({sinx}−{x}\right)} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }×{x}^{\mathrm{2}} }{\:\left(\mathrm{1}−{x}\right)×\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)×−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} }\: \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\checkmark \\ $$

Answered by cortano12 last updated on 18/Jul/23

lim_(x→0) (((√((1+x)/(1−x)))−(1+x))/(((sin x)/x)−1)) = lim_(x→0) (((√(1+x))−(1+x)(√(1−x)))/( (√(1−x)) (((sin x−x)/x)))) = lim_(x→0) (((√(1+x)) (1−(√(1−x^2 )) ))/( (√(1−x)) (((sin x−x)/x)))) = lim_(x→0) ((√(1−x))/( (√(1−x)))) .lim_(x→0) (x^3 /((1+(√(1−x^2 )))(sin x−x))) = (1/2).(1/(lim_(x→0) (((sin x−x)/x^3 )))) = determinant ((((1/2).((1/(−1/6)))= −3)))

$$\:\: \underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}−\left(\mathrm{1}+{x}\right)}{\frac{\mathrm{sin}\:{x}}{{x}}−\mathrm{1}} \\ $$$$\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+{x}}−\left(\mathrm{1}+{x}\right)\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}−{x}}\:\left(\frac{\mathrm{sin}\:{x}−{x}}{{x}}\right)} \\ $$$$\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+{x}}\:\left(\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\right)}{\:\sqrt{\mathrm{1}−{x}}\:\left(\frac{\mathrm{sin}\:{x}−{x}}{{x}}\right)} \\ $$$$\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}−{x}}}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\left(\mathrm{sin}\:{x}−{x}\right)} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:{x}−{x}}{{x}^{\mathrm{3}} }\right)} \\ $$$$\:\:=\:\begin{array}{|c|}{\frac{\mathrm{1}}{\mathrm{2}}.\left(\frac{\mathrm{1}}{−\mathrm{1}/\mathrm{6}}\right)=\:−\mathrm{3}}\\\hline\end{array} \\ $$$$\:\: \\ $$$$ \\ $$

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