STD 11 - 12. limits — JEE STD 11 Science — Question

$ = {e^{\frac{1}{x}\,\left( {x\, - \,\frac{{{x^2}}}{2}\, + \,\frac{{{x^3}}}{3}\, - \,\frac{{{x^4}}}{4}\, + ....} \right)}}$$ = {e^{\left( {1\, - \,\frac{x}{2}\, + \,\frac{{{x^2}}}{3}\, - \,\frac{{{x^3}}}{4}\, + \,....} \right)}}$

$ = e.{e^{\left( {\, - \,\frac{x}{2}\, + \,\frac{{{x^2}}}{3}\, - \,\frac{{{x^3}}}{4} + ....} \right)}}$

$ = e\left[ {\frac{{\left( { - \frac{x}{2} + \frac{{{x^2}}}{3} - \frac{{{x^3}}}{4} + ...} \right)}}{{1!}} + \frac{{{{\left( { - \frac{x}{2} + \frac{{{x^2}}}{3} - \frac{{{x^3}}}{4} + ...} \right)}^2}}}{{2!}} + ...} \right]$

$ = \left[ {e - \frac{{ex}}{2} + \frac{{11e}}{{24}}{x^2} + ... + ...} \right]$

$\therefore$ $\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/x}} - e}}{x}$$ = \mathop {\lim }\limits_{x \to 0} \,\left[ {\frac{{e - \frac{{ex}}{2} - \frac{{11e}}{{24}}{x^2} + ...e}}{x}} \right]$

==> $\mathop {\lim }\limits_{x \to 0} \,\left( { - \frac{e}{2} - \frac{{11e}}{{24}}x + ...} \right)$$ = - \frac{e}{2}$.