In some references regarding sign consistency of Lasso such as
Zhao, Peng; Yu, Bin, On model selection consistency of Lasso, J. Mach. Learn. Res. 7, 2541-2563 (2006). ZBL1222.62008.
there is a condition such that
$\frac{\lambda_n}{n} \rightarrow 0, \frac{\lambda_n}{\sqrt{n}} \rightarrow \infty$
(In the above paper $\sqrt{n}$ is indeed $n^\frac{1+c}{2}$ for $0 \leq c < 1$). I understand $\frac{\lambda_n}{n} \rightarrow 0$ as $n$ goes infinity, but how does $\frac{\lambda_n}{\sqrt{n}} \rightarrow \infty$ hold at the same time? Thanks very much in advance!