STD 11 - 12. limits — JEE कक्षा 11 साइन्स — Question

$\mathop {\lim }\limits_{x \to 0} \,\frac{{\cos x - \frac{1}{{1 - x}}}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \sin x - \frac{1}{{{{(1 - x)}^2}}}}}{2} = - \frac{1}{2}$.

वैकल्पिक : $\mathop {\lim }\limits_{x \to 0} \,\frac{{\sin x + \log \,(1 - x)}}{{{x^2}}}$

$ = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{\left( {x - \frac{{{x^3}}}{{3\,\,!}} + \frac{{{x^5}}}{{5\,\,!}} - ...} \right)}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \,\,\frac{{\left( { - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} - ...} \right)}}{{{x^2}}}$

$\left( \because \sin x=x-\frac{{{x}^{3}}}{3\,!}+\frac{{{x}^{5}}}{5\,!}-.. \right.$ व

$\left. {\log \,(1 - x) = - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - ..} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{\frac{{ - {x^2}}}{2} - {x^3}\left( {\frac{1}{{3\,\,!}} + \frac{1}{3}} \right) - \frac{{{x^4}}}{4}...}}{{{x^2}}} = - \frac{1}{2}.$