Evaluate $\lim\limits_{x \to 0} \; \dfrac{e^x \sin x - x -x^2}{x^2 + x \log \left(1-x\right)}$
$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{e^x \sin x -x -x^2}{x^2 + x \log \left(1-x\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left\{e^x \sin x -x - x^2\right\}}{\dfrac{d}{dx}\left\{x^2 + x \log \left(1-x\right)\right\}} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{e^x \cos x + e^x \sin x -1 - 2x}{2x-\dfrac{x}{1-x}+\log \left(1-x\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left\{e^x \cos x + e^x \sin x -1 -2x\right\}}{\dfrac{d}{dx}\left\{2x - \dfrac{x}{1-x}+ \log \left(1-x\right) \right\}} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{e^x \cos x -e^x \sin x + e^x \sin x + e^x \cos x -2}{2 - \dfrac{1-x+x}{\left(1-x\right)^2}- \dfrac{1}{1-x}} & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{2e^x \cos x -2}{2 - \dfrac{1}{\left(1-x\right)^2}-\dfrac{1}{1-x}} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left\{2e^x \cos x -2\right\}}{\dfrac{d}{dx}\left\{2-\dfrac{1}{\left(1-x\right)^2}-\dfrac{1}{1-x}\right\}} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{2\left(e^x \cos x -e^x \sin x\right)}{\dfrac{-2}{\left(1-x\right)^3}- \dfrac{1}{\left(1-x\right)^2} } & \\ & & \\ & = \dfrac{2\times \left(1-0\right)}{-2-1} & \\ & & \\ & = - \dfrac{2}{3} \end{aligned}$