Two sensors give readings that grow without bound near a point and far from it…

Step 1. The first difference. $\dfrac{1}{x - 1} - \dfrac{2}{x^2 - 1} = \dfrac{(x + 1) - 2}{x^2 - 1} = \dfrac{x - 1}{(x - 1)(x + 1)} = \dfrac{1}{x + 1}$. As $x \to 1$ it tends to $\dfrac{1}{2} = 0.5$.

Step 2. The second difference. The conjugate gives $\dfrac{(x^2 + 4x) - (x^2 - 2x)}{\sqrt{x^2 + 4x} + \sqrt{x^2 - 2x}} = \dfrac{6x}{\sqrt{x^2 + 4x} + \sqrt{x^2 - 2x}}$.

Step 3. Divide by $x$. For $x > 0$ this is $\dfrac{6}{\sqrt{1 + 4/x} + \sqrt{1 - 2/x}} \to \dfrac{6}{1 + 1} = 3$.