Answer
$\dfrac{1}{3}$
Work Step by Step
Our aim is to find the limit by using a Taylor's series.
$L=\lim\limits_{x \to 0} \dfrac{\ln (1+x)-x+\dfrac{x^2}{2}}{x^3}$
Recall the series for $\ln (1+x)=\Sigma_{k =1}^{\infty} (-1)^{k+1} \dfrac{(x)^{k}}{k}$
Now, $L=\lim\limits_{x \to 0} \dfrac{\Sigma_{k =1}^{\infty} (-1)^{k+1} \dfrac{(x)^{k}}{k}-x+\dfrac{x^2}{2}}{x^3}\\=\lim\limits_{x \to 0}[\Sigma_{k =1}^{\infty} (-1)^{k+1} \dfrac{(x)^{k-3}}{k}-\dfrac{1}{x^2}+\dfrac{1}{2x}]\\=\lim\limits_{x \to 0} [ (-1)^{1+1} \dfrac{(x)^{1-3}}{1}+(-1)^{2+1} \dfrac{(x)^{2-3}}{2}+.......)-\dfrac{1}{x^2}+\dfrac{1}{2x}]\\=\lim\limits_{x \to 0} [\dfrac{1}{3}-\dfrac{x}{4}+\dfrac{x^2}{5}+......]\\=\dfrac{1}{3}-\dfrac{0}{4}+\dfrac{(0)^2}{5}+......\\=\dfrac{1}{3}$
