Let R be the reals. f **:** [1, ∞) → R is differentiable and satisfies f '(x) = 1/(x^{2} + f(x)^{2}) and f(1) = 1. Show that as x → ∞, f(x) tends to a limit which is less than 1 + π/4.

**Solution**

Clearly f '(x) is always positive. But f(1) = 1, so f(x) > 1 for all x. Hence f '(x) < 1/(x^{2} + 1) for all x. Hence f(x) = 1 + ∫_{1}^{x} f '(t) dt < 1 + 1 + ∫_{1}^{x} 1/(1 + t^{2}) dt = 1 + (tan^{-1}t)|_{1}^{x} = 1 + tan^{-1}x - π/4 < 1 + π/2 - π/4 = 1 + π/4. Since f '(x) is positive, f(x) is monotone increasing. It is bounded above by 1 + π/4, so it must tend to a limit less than 1 + π/4.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002