Prove that $u^{-1}\int_{-u}^u(1-\varphi(t))dt\to 0$ as $u\to 0$

Question

How to prove that $u^{-1}\int_{-u}^u(1-\varphi(t))dt\to 0$ as $u\to 0$

$\varphi(t)$ is the characteristic function of a random variable, if $u\to 0$ then $\varphi(t)\to 1$, so one gets $\frac00$ can we use then l'Hopital, I mean $\lim\limits_{u\to 0}u^{-1}\int_{-u}^u(1-\varphi(t))dt=\lim\limits_{u\to 0}\frac{-\varphi(u)+\varphi(-u)}{1}=0$

Answer 1

Take $\varepsilon >0$. There exists $\delta >0$ such that $|t|<\delta$ implies $|1-\varphi(t)|<\varepsilon$. Then if $0<u<\delta$, $$\frac{1}u \int_{(-u,u)} |1-\varphi(t)|dt <\frac{1}{u}2u \varepsilon=2\varepsilon$$