IMC 2005 · Day 1 · P5

Let $g(x) = f'(x) + x f(x)$; then $f''(x) + 2x f'(x) + (x^2 + 1) f(x) = g'(x) + x g(x)$.

We prove that if $h$ is a continuously differentiable function such that $h'(x) + x h(x)$ is bounded then $\lim\limits_{\infty} h = 0$. Applying this lemma for $h = g$ then for $h = f$, the statement follows.

Let $M$ be an upper bound for $|h'(x) + x h(x)|$ and let $p(x) = h(x) e^{x^2/2}$. (The function $e^{-x^2/2}$ is a solution of the differential equation $u'(x) + x u(x) = 0$.) Then $$|p'(x)| = |h'(x) + x h(x)| e^{x^2/2} \le M e^{x^2/2}$$ and $$|h(x)| = \left| \frac{p(x)}{e^{x^2/2}} \right| = \left| \frac{p(0) + \int_0^x p'}{e^{x^2/2}} \right| \le \frac{|p(0)| + M \int_0^x e^{x^2/2}\,dx}{e^{x^2/2}}.$$ Since $\lim\limits_{x \to \infty} e^{x^2/2} = \infty$ and $\lim\limits_{x \to \infty} \frac{\int_0^x e^{x^2/2}\,dx}{e^{x^2/2}} = 0$ (by L'Hospital's rule), this implies $\lim\limits_{x \to \infty} h(x) = 0$.

Apply L'Hospital rule twice on the fraction $\frac{f(x) e^{x^2/2}}{e^{x^2/2}}$. (Note that L'Hospital rule is valid if the denominator converges to infinity, without any assumption on the numerator.) $$\begin{align*} \lim_{x \to \infty} f(x) &= \lim_{x \to \infty} \frac{f(x) e^{x^2/2}}{e^{x^2/2}} = \lim_{x \to \infty} \frac{(f'(x) + x f(x)) e^{x^2/2}}{x e^{x^2/2}} = \lim_{x \to \infty} \frac{(f''(x) + 2x f'(x) + (x^2+1) f(x)) e^{x^2/2}} {(x^2 + 1) e^{x^2/2}} = \\ &= \lim_{x \to \infty} \frac{f''(x) + 2x f'(x) + (x^2+1) f(x)}{x^2 + 1} = 0. \end{align*}$$