IMC 2001 · Day 1 · P6

Let $0 < \varepsilon < A$ be an arbitrary real number. If $x$ is sufficiently large then $f(x) > 0$, $g(x) > 0$, $|a(x) - A| < \varepsilon$, $|b(x) - B| < \varepsilon$ and $$\begin{align*} B - \varepsilon < b(x) &= \frac{f'(x)}{g'(x)} + a(x) \frac{f(x)}{g(x)} < \frac{f'(x)}{g'(x)} + (A + \varepsilon) \frac{f(x)}{g(x)} < \tag{1}\\ &< \frac{(A+\varepsilon)(A+1)}{A} \cdot \frac{f'(x) \bigl( g(x) \bigr)^A + A \cdot f(x) \cdot \bigl( g(x) \bigr)^{A-1} \cdot g'(x)} {(A+1) \cdot \bigl( g(x) \bigr)^A \cdot g'(x)} = \\ &= \frac{(A+\varepsilon)(A+1)}{A} \cdot \frac{\Bigl( f(x) \cdot \bigl( g(x) \bigr)^A \Bigr)'} {\Bigl( \bigl( g(x) \bigr)^{A+1} \Bigr)'}, \end{align*}$$ thus $$\tag{2} \frac{\Bigl( f(x) \cdot \bigl( g(x) \bigr)^A \Bigr)'} {\Bigl( \bigl( g(x) \bigr)^{A+1} \Bigr)'} > \frac{A (B - \varepsilon)}{(A + \varepsilon)(A + 1)}.$$ It can be similarly obtained that, for sufficiently large $x$, $$\tag{3} \frac{\Bigl( f(x) \cdot \bigl( g(x) \bigr)^A \Bigr)'} {\Bigl( \bigl( g(x) \bigr)^{A+1} \Bigr)'} < \frac{A (B + \varepsilon)}{(A - \varepsilon)(A + 1)}.$$ From $\varepsilon \to 0$, we have $$\lim_{x \to \infty} \frac{\Bigl( f(x) \cdot \bigl( g(x) \bigr)^A \Bigr)'} {\Bigl( \bigl( g(x) \bigr)^{A+1} \Bigr)'} = \frac{B}{A + 1}.$$ By l'Hospital's rule this implies $$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f(x) \cdot \bigl( g(x) \bigr)^A}{\bigl( g(x) \bigr)^{A+1}} = \frac{B}{A + 1}.$$