Processing math: 100%

Wednesday, June 8, 2016

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}

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