Comparing Fractions

$A = \frac {1}{x + 1} + \frac {1}{x - 1} = \frac {2x} {x^2 - 1}$

$B = \frac {1}{x + 2} + \frac {1}{x - 2} = \frac {2x}{x^2 - 4}$

As we can see, the numerators are the same (and since x > 100, both the numerators and the denominators are positive), therefore the fraction with the smaller denominator will be the bigger number:

$x^2 - 4 < x^2 - 1 \iff -4 < -1$ , therefore $\boxed {A < B}$ .

Consider the following:

$\begin{aligned} \frac AB & = \frac {\frac 1{x-1} + \frac 1{x+1}}{\frac 1{x-2} + \frac 1{x+2}} = \frac {\frac {2x}{x^2-1}}{\frac {2x}{x^2-4}} = \frac {x^2-4}{x^2-1} = 1 - \frac 3{x^2-1} < 1 \end{aligned}$ for $|x| > 1$ .

Therefore, $\boxed{B} > A$ for $x > 100$ .

From a graphical perspective, the magnitude of the slope of $\frac{1}{x}$ gets smaller as $x$ gets larger. This means that:

$\frac{1}{x-2}$ - $\frac{1}{x-1}$ > $\frac{1}{x+1}$ - $\frac{1}{x+2}$

Re-arranging gives:

$\frac{1}{x-2}$ + $\frac{1}{x+2}$ > $\frac{1}{x-1}$ + $\frac{1}{x+1}$