Suppose that $f(x)$ is a continuous function on $(0,1]$, and moreover that $$\int_{0}^{1}f(x)dx < \infty$$. In this setting, $f$ has an integrable singularity at $0$
My Question: $$f(x)=o(\frac{1}{x}) \text{, as } x \to 0^{+}$$?
Suppose that $f(x)$ is a continuous function on $(0,1]$, and moreover that $$\int_{0}^{1}f(x)dx < \infty$$. In this setting, $f$ has an integrable singularity at $0$
My Question: $$f(x)=o(\frac{1}{x}) \text{, as } x \to 0^{+}$$?
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