Mathematical Reflections 2016, Issue 2 - Problem U372

Problem:
Let $\alpha,\beta>0$ be real numbers and let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function such that $f(x) \neq 0$ for all $x$ in a neighborhood $U$ of $0$. Evaluate $$\lim_{x \to 0^+} \dfrac{\int_0^{\alpha x} t^{\alpha}f(t) \ dt}{\int_0^{\beta x} t^{\beta}f(t) \ dt}.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Observe that $\lim_{x \to 0+} \int_0^{\alpha x} t^{\alpha}f(t) \ dt=\lim_{x \to 0^+} \int_0^{\beta x} t^{\beta}f(t) \ dt=0$. Let $g(t)=t^{\alpha}f(t)$ and $h(t)=t^{\beta} f(t)$. We have $$\dfrac{d}{dx} \int_0^{\alpha x} g(t) \ dt=g(\alpha x)\alpha=(\alpha x)^{\alpha}f(\alpha x)\alpha$$ and $$\dfrac{d}{dx} \int_0^{\beta x} h(t) \ dt=h(\beta x)\beta=(\beta x)^{\beta}f(\beta x)\beta.$$ Clearly, $\dfrac{d}{dx} \int_0^{\beta x} h(t) \ dt \neq 0$ for all $x$ in a neighborhood $U$ of $0$ ($x \neq 0$). We have $$\lim_{x \to 0^+}\dfrac{\dfrac{d}{dx} \int_0^{\alpha x} t^{\alpha}f(t) \ dt}{\dfrac{d}{dx} \int_0^{\beta x} t^{\beta}f(t) \ dt}=\begin{cases} 0 & \textrm{ if } \alpha>\beta \\ 1 & \textrm{ if } \alpha=\beta \\ +\infty & \textrm{ if } \alpha<\beta.  \end{cases}$$
By L'H\^{o}pital's Rule we conclude that
$$\lim_{x \to 0^+} \dfrac{\int_0^{\alpha x} t^{\alpha}f(t) \ dt}{\int_0^{\beta x} t^{\beta}f(t) \ dt}=\begin{cases} 0 & \textrm{ if } \alpha>\beta \\ 1 & \textrm{ if } \alpha=\beta \\ +\infty & \textrm{ if } \alpha<\beta.  \end{cases}$$