Question Number 194812 by sonukgindia last updated on 16/Jul/23
Answered by MM42 last updated on 16/Jul/23
$${A}={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{1}−{x}}−\left(\mathrm{1}+{x}\right)}{\:\frac{{sinx}−{x}}{{x}}} \\ $$$${A}={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\left(\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{\:\left(\mathrm{1}−{x}\right)\left({sinx}−{x}\right)} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }×{x}^{\mathrm{2}} }{\:\left(\mathrm{1}−{x}\right)×\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)×−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} }\: \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\checkmark \\ $$
Answered by cortano12 last updated on 18/Jul/23
$$\:\:\:\underline{\underbrace{\lessdot}\cancel{} \underbrace{\vdots\boldsymbol{\iota}\spadesuit\spadesuit\left[}}\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}−\left(\mathrm{1}+{x}\right)\right. \\ $$