Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

   ⋖ ⋮𝛊♠♠[(√((1+x)/(1−x)))−(1+x)

$$\:\:\:\underline{\underbrace{\lessdot}\cancel{} \underbrace{\vdots\boldsymbol{\iota}\spadesuit\spadesuit\left[}}\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}−\left(\mathrm{1}+{x}\right)\right. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com