Given $0<r<1.$ Define the function $f: [0,+\infty)\to \mathbb{R}$ as $$f(x) := x\int_0^\infty t^{-r}(t+x)^{-1} dt,\quad x>0 \quad \text{and}\quad f(0) :=0. $$ I want to prove that $f$ is continuous on $[0,+\infty).$
I started as follows: taking $\epsilon >0,$ we can prove the continuity on $[\epsilon, +\infty)$ using this theorem where it suffices to notice that for the third assumption we have $$t^{-r}(t+x)^{-1} \leq \frac 1\epsilon\chi_{(0,1)}(t)t^{-r} +\chi_{(1,+\infty)}(t)t^{-(r+1)}.$$ This implies the continuity of $x\mapsto \int_0^\infty t^{-r}(t+x)^{-1} dt$ on $[\epsilon, +\infty).$ Since $\epsilon >0$ is arbitrary, we conclude that $x\mapsto \int_0^\infty t^{-r}(t+x)^{-1} dt$ is continuous on $(0,+\infty).$ Hence $f$ is continuous on $(0,+\infty).$
Then it remains to prove that $\lim_{x\to 0^+} f(x) = 0.$ Thank you for any hint.