Answer
$\dfrac{1}{2}$
Work Step by Step
Our aim is to find the limit by using a Taylor's series.
$L=\lim\limits_{x \to 0} \dfrac{-x-\ln (1-x)}{x^2}$
Recall the series for $\ln (1-x)=-\Sigma_{k =1}^{\infty} (-1)^k \dfrac{(-x)^{k}}{k}$
Now, $L=\lim\limits_{x \to 0} \dfrac{-x-\ln (1-x)}{x^2} \\=\lim\limits_{x \to 0} \dfrac{-x--\Sigma_{k =1}^{\infty} (-1)^k \dfrac{(-x)^{k}}{k}}{x^2} \\=\lim\limits_{x \to 0}(\dfrac{-x}{x^2}+\Sigma_{k =1}^{\infty} \dfrac{x^{k-2}}{k}) \\=\lim\limits_{x \to 0} (\dfrac{-1}{x}+\dfrac{1}{x}+\Sigma_{k =2}^{\infty} \dfrac{x^{k-2}}{k}) \\=\lim\limits_{x \to 0} [\Sigma_{k =2}^{\infty} \dfrac{x^{k-2}}{k}]\\=\lim\limits_{x \to 0} [ \dfrac{x^{2-2}}{2}+\dfrac{x^{3-2}}{3}+\dfrac{x^{4-2}}{4}+......) \\=\dfrac{1}{2}+\dfrac{0}{3}+\dfrac{0^2}{4}+........\\=\dfrac{1}{2}$
