$\newcommand{\R}{\mathbb R}$For natural $n$, $a\in\R^n$, and real $t>0$, let  
\begin{equation*}
	K:=K_{n,t}(a):=\inf_{x\in\R^n}(\|a-x\|_2+t\|x\|_1),
\end{equation*}
\begin{equation*}
	M:=M_{n,t}(a):=\min(\|a\|_2,t\|a\|_1),   
\end{equation*}
and (for nonzero $a$) 
\begin{equation*}
	R:=R_{n,t}(a):=\frac LM,   
\end{equation*}
where $\|x\|_p:=(\sum_1^n|x_i|^p)^{1/p}$ for $x=(x_1,\dots,x_n)\in\R^n$. 

So, the function $K_{n,t}$ is a norm on $\R^n$, which is the [infimal convolution][1] of the norms $\|\cdot\|_2$ and $t\|\cdot\|_1$. The function $M_{n,t}$ is a norm only for $t\ge1$ (and then $M_{n,t}=\|\cdot\|_2$) and $t\le1/\sqrt n$ (and then $M_{n,t}=t\|\cdot\|_1$). 

Clearly, $K\le M$. 
It was [previously asked][2] whether, for each $t>0$,  
\begin{equation*}
	\inf_{a\in\R^n\setminus\{0\}}R_{n,t}(a)\to0
\end{equation*}
as $n\to\infty$. 

It was then shown that this is [not true for $t=1$][3] and also [not true for any real $t>0$][4], because 
$$\frac KM\ge\min(1,t).$$ 

It was [further asked][5] if 
\begin{equation*}
	\inf_{a\in\R^n\setminus\{0\}}R_{n,t_n}(a)\to0
\end{equation*}
as $n\to\infty$ assuming that $t_n\to0$. 

A somewhat surprising answer to this question will be given below.  

  [1]: https://en.wikipedia.org/wiki/Convex_conjugate#Infimal_convolution 
  [2]: https://mathoverflow.net/q/432168/36721
  [3]: https://mathoverflow.net/questions/432168/fix-positive-t-construct-a-n-in-mathbb-rn-such-that-inf-x-x-a-n-2#comment1112378_432168 
  [4]: https://mathoverflow.net/a/432184/36721 
  [5]: https://mathoverflow.net/questions/432168/fix-positive-t-construct-a-n-in-mathbb-rn-such-that-inf-x-x-a-n-2#comment1112883_432184