Question

If $\underset{x\to 0}{lim}\frac{{(1+x)}^{1/x}-e+\frac{ex}{2}}{x^2}=\frac{ae}{24}$, the find $a$.

Answer 1

Let, $y={(1+x)}^{1/x}$
Then,
$\log y=\frac{\log (1+x)}{x}=\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots )}{x}=(1-\frac{x}{2}+\frac{x^2}{3}-\cdots )$
$\Rightarrow y=e^{1-(\frac{x}{2}-\frac{x^2}{3}+.....)}=e.e^{-(\frac{x}{2}-\frac{x^2}{3}+....)}$
$=e[1-(\frac{x}{2}-\frac{x^2}{3})+\frac{1}{2!}{(\frac{x}{2}-\frac{x^2}{3})}^2+\cdots ]$
$=e[1-\frac{x}{2}+x^2(\frac{1}{3}+\frac{1}{8})+x^3+\cdots ]$
$\therefore y-e+\frac{ex}{2}=e{x}^2.\frac{11}{24}+e{x}^3\left(\right)+\cdots$
$\therefore \underset{x\to 0}{Lt}\frac{{(1+x)}^{1/x}-e+\frac{ex}{2}}{x^2}=\frac{11e}{24}.$
$\Rightarrow a=11$