Mathematics - Limits Question with Solution | TestHub

$\underset{x\to 0^{+}}{\lim }\frac{{\left(1-x\right)}^{{\displaystyle \frac{1}{x}}}}{a{x}^{a-1}}\frac{d}{dx}\left[\frac{\ln \left(1-x\right)}{x}\right]=\underset{x\to 0^{+}}{\lim }\frac{{\left(1-x\right)}^{\frac{1}{x}}}{a{x}^{a-1}}\left[\frac{1}{\left(x\right)\left(x-1\right)}-\frac{\ln \left(1-x\right)}{x^2}\right]$

$=\underset{x\to 0^{+}}{\lim }\frac{{\left(1-x\right)}^{{\displaystyle \frac{1}{x}}}}{a{x}^{a-1}}\left[\frac{x-\left(x-1\right)\ln \left(1-x\right)}{x^2\left(x-1\right)}\right]$

$=\underset{x\to 0^{+}}{\lim }\frac{{\left(1-x\right)}^{\frac{1}{x}}\left\{x+\left(x-1\right)\left(x+\frac{x^2}{2}+\frac{x^3}{3}........\right)\right\}}{-a\left(1-x\right)x^{a+1}}$

$=\underset{x\to 0^{+}}{\lim }\frac{{\left(1-x\right)}^{\frac{1}{x}}\left\{\frac{x^2}{2}+x^3\left(\frac{1}{2}-\frac{1}{3}\right)+x^4\left(\frac{1}{3}-\frac{1}{4}\right)+\dots \right\}}{-a\left(1-x\right)x^{a+1}}$

Which is real number if and only if $a+1=2\Rightarrow a=1$