$\newcommand{\la}{\lambda}$Write 
\begin{equation*}
	Y=\sqrt{V_k+\la^2},
\end{equation*}
where $V_k$ is a random variable (r.v.) with the chi-squared distribution with $k$ degrees of freedom. 

So, by the law of large numbers and the condition $\la\sim L\sqrt k$ (as $k\to\infty$),
\begin{equation*}
	\frac{Y}{\sqrt k}=\sqrt{\frac{V_k}k+\frac{\la^2}k}\to\sqrt{1+L^2} 
\end{equation*}
in probability and hence in distribution. 
Also, by the central limit theorem (or, more precisely, by, say, the Berry--Esseen inequality), 
\begin{equation*}
	\frac{Y^2-(k+\la^2)}{\sqrt{2k}}\to Z\sim N(0,1) 
\end{equation*}
in distribution. 
So, by Slutsky's theorem,
\begin{equation*}
	Y-\sqrt{k+\la^2}=R_k:=\frac{(Y^2-(k+\la^2))/\sqrt{2k}}{(Y+\sqrt{k+\la^2})/\sqrt{2k}}
	\to\frac Z{\sqrt{2(1+L^2)}} 
\end{equation*}
in distribution, where again $Z\sim N(0,1)$. 

Also, by (say) the Rosenthal inequality, 
\begin{equation*}
	ER_k^4\le E\Big(\frac{(Y^2-(k+\la^2))/\sqrt{2k}}{(\sqrt{k+\la^2})/\sqrt{2k}}\Big)^4=O(1). 
\end{equation*}
So, by uniform integrability, 
\begin{equation*}
	E(Y-\sqrt{k+\la^2})\to E\frac Z{\sqrt{2(1+L^2)}}=0 
\end{equation*}
and 
\begin{equation*}
	E(Y-\sqrt{k+\la^2})^2\to E\Big(\frac Z{\sqrt{2(1+L^2)}}\Big)^2=\frac1{2(1+L^2)}.  
\end{equation*}
Thus, 
\begin{equation*}
	EY=\sqrt{k+\la^2}+o(1) \tag{1}\label{1}
\end{equation*}
and 
\begin{equation*}
Var\, Y=	E(Y-\sqrt{k+\la^2})^2-(E(Y-\sqrt{k+\la^2}))^2\to\frac1{2(1+L^2)}.  \tag{2}\label{2}
\end{equation*}

---

For an illustration of \eqref{1} and \eqref{2}, shown below are (parts of) the (connected) graphs $\{(k,EY-\sqrt{k+\la^2})\colon k\in\{1,\dots,50\}\}$ (left) and $\{(k,Var\,Y-\frac1{2(1+L^2)}\colon k\in\{1,\dots,50\}\}$ (right) for $\la=L\sqrt k$ with $L=0$ (red), $L=1$ (green), and $L=2$ (blue): 

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/MtzbD.png