$\lim_{x \to \infty} \frac{1}{x^{1-\frac{1}{p}}}\int_0^xf=0$ where $f \in L^p$ $1<p<\infty$.
Proof: We approximate $f$ with the functions $f\chi_{[-n,n]}$. This sequence of functions converges in $L^p$ norm to $f$. The statement is clearly true for functions with compact support, thus $|\frac{1}{x^{1-\frac{1}{p}}}\int_0^xf|\leq |\frac{1}{x^{1-\frac{1}{p}}}\int_0^x(f-f_n)|+|\frac{1}{x^{1-\frac{1}{p}}}\int_0^xf_n|<\epsilon$ by picking $n$ big enough so that $||f-f_n||<\epsilon/2$ and $x$ big enough so that for that $n$, $|\frac{1}{x^{1-\frac{1}{p}}}\int_0^xf_n|<\epsilon/2$ This concludes the proof.