Let $\Phi(u)$ be the standard normal distribution function. We want to consider the following expression $\|\Phi(u/\sigma\_l)-\Phi(u)\|\_{\infty}$, where $\sigma\_l\xrightarrow{l\rightarrow\infty}1$. For convenience let's say $\sigma\_l=1-\frac{1}{l}$.

Is it possible to approximate the expression above in the following form:
\begin{align\*}\|\Phi(u/\sigma\_l)-\Phi(u)\|\_{\infty}\leq f(l)\end{align\*} where $f:\mathbb{N}\rightarrow\mathbb{R}$ with $\lim\_{l\rightarrow\infty}f(l)=0$?
Thanks a lot!