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