Limits of Functions!

The function $g(x) = f(x)+x$ satisfies the differential equation: $g'(x) = 1 + \frac{1 - (g(x)/x - 1)^2}{1 + x^2(g(x)/x - 1)^2}.$ This implies that $g'(x) > 0$ for all $x > 1$ , so the limit $L_1 = \lim_{x \to +\infty} g(x)$ exists. In addition, we cannot have $L_1 < +\infty$ , or else we would have $\lim_{x \to +\infty} g'(x) = 0$ whereas the differential equation forces this limit to be 1. Hence $g(x) \to +\infty$ as $x \to +\infty$ .

Similarly, the function $h(x) = -f(x) + x$ satisfies the differential equation $h'(x) = 1 - \frac{1 - (h(x)/x - 1)^2}{1 + x^2(h(x)/x - 1)^2}.$ This implies that $h'(x) \geq 0$ for all $x$ , so the limit $L_2 = \lim_{x \to +\infty} h(x)$ exists. In addition, we cannot have $L_2 < +\infty$ , or else we would have $\lim_{x \to +\infty} h'(x) = 0$ whereas the differential equation forces this limit to be 1. Hence $h(x) \to +\infty$ as $x \to +\infty$ .

For some $x_1 > 1$ , we must have $g(x), h(x) > 0$ for all $x \geq x_1$ . For $x \geq x_1$ , we have $|f(x)| < x$ and hence $f'(x) > 0$ , so the limit $L = \lim_{x \to +\infty} f(x)$ exists. Once again, we cannot have $L < +\infty$ , or else we would have $\lim_{x \to +\infty} f'(x) = 0$ whereas the original differential equation (e.g., in the form given in the first solution) forces this limit to be $1/(1 + L^2) > 0$ . Hence $f(x) \to +\infty$ as $x \to \infty$ , as desired.