MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/x}} - e + \frac{1}{2}ex}}{{{x^2}}} = . . .$
- ✓$\frac{{11e}}{{24}}$
- B$\frac{{ - 11e}}{{24}}$
- C$\frac{e}{{24}}$
- Dએકપણ નહી.
$ = {e^{1 - \frac{x}{2} + \frac{{{x^2}}}{3}\, - ......}} = e\,{e^{ - \,\frac{x}{2} + \frac{{{x^2}}}{3}\, - \,.....}}$
$ = e\,\left[ {1 + \left( { - \frac{x}{2} + \frac{{{x^2}}}{3} - .....} \right) + \frac{1}{{2\,\,!}}\,{{\left( { - \frac{x}{2} + \frac{{{x^2}}}{3}\, - \,.....} \right)}^2} + ...} \right]$
$ = e\,\left[ {1 - \frac{x}{2} + \frac{{11}}{{24}}{x^2} - ....} \right]$
$\therefore \,\,\,\mathop {\lim }\limits_{x \to 0} \,\,\frac{{{{(1 + x)}^{1/x}} - e + \frac{{ex}}{2}}}{{{x^2}}} = \frac{{11e}}{{24}}$.
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