Showing $u_{\lambda}(x):= \left(\frac{\lambda}{{\lambda}^{2}+|x|^2}\right)^{\frac{n-2}{2}}$ is not sequentially compact in $L^{2^{*}}$ 
Assume $n\geq 3$, consider $u(x):= (1+|x|^2)^\frac{2-n}{2}$ ($x \in \mathbb{R}^{n}$) and for each $\lambda>0$, define $$u_{\lambda}(x):= \left(\frac{\lambda}{{\lambda}^{2}+|x|^2}\right)^{\frac{n-2}{2}}$$. Show that $\{u_\lambda\}_{\lambda>0}$ is not precompact in $L^{2^{*}}(\mathbb{R}^n)$, where $2^{*}=\frac{2n}{n-2}$.

Here, precompactness means for any sequence in $\{u_\lambda\}_{\lambda>0}$, there exists a convergent subsequence in $L^{2^{*}}(\mathbb{R}^n)$.
As of now, I can show that $\lVert u_\lambda \rVert_{L^{2^*}(\mathbb{R}^{n})}=\lVert u\rVert_{L^{2^*}(\mathbb{R}^{n})}$ and $\lVert Du_\lambda \rVert_{L^2(\mathbb{R}^{n})}=\lVert Du\rVert_{L^2(\mathbb{R}^{n})}$. I also notice that for any fixed $x \in \mathbb{R}^n$, $u_\lambda (x) \rightarrow 0$ as $\lambda \rightarrow \infty$. Hence, I attempt to prove by contradiction. Suppose the family is precompact then pick a sequence $\{u_{\lambda_{k}}\}_{k \in \mathbb{N}}$, where $\lambda_{k}\rightarrow \infty$ as $k \rightarrow \infty$. By precompactness, say $\{u_{\lambda_{k_j}}\}_{j \in \mathbb{N}}$ is a subsequence converging to (say) $v$ in $L^{2^{*}}(\mathbb{R}^n)$. From triangle inequality and the equalities mentioned above, we have  $\lVert u\rVert_{L^{2^*}(\mathbb{R}^{n})}= \lVert v\rVert_{L^{2^*}(\mathbb{R}^{n})}$ at which point I am stuck. I think one should also somehow use the Gagliardo-Nirenberg inequality, which provides a constant $C=C(2,n)$ such that $\lVert u\rVert_{L^{2^*}(\mathbb{R}^{n})} \leq C \lVert Du\rVert_{L^2(\mathbb{R}^{n})}$. 
Any help or suggestion is greatly appreciated. 
 A: If $u_{\lambda_{k_j}}$ converges to some $f\in L^{2^*}$ as $j\to \infty$, then by taking a further subsequence if necessary, $u_{k_j}$ converges pointwisely to $f$. This implies that $f = 0$ a.e.. But this is not possible as
$$C= \| u_\lambda\|_{2^*} =  \| u_\lambda - f\|_{2^*}$$
A: Assume that there is a sequence  $\lambda_j$ and a function $v$ such that 
$$\|u_{\lambda_j} -v\|_{2^*}\to0$$
By the converse of Lebesgue dominated convergence theorem there exists a subsequence $\lambda_{j_k}$ such that 
$$u_{\lambda_ {j_k}}(x)\to v(x)~~~~a.e$$
Patently this gives $v(x)=0.$ since $u_{\lambda} \to0$ as ${\lambda}\to\infty$. However, this is contradictory since $$0= \lim_{j\to\infty}\|u_{\lambda_j}-v\|_{2^*}^{2^*}= \lim_{j\to\infty}\|u_{\lambda_j}\|_{2^*}^{2^*} =c_n\neq0 $$
indeed,  enforcing the change of variables $x= \lambda z$ we have
$$\|u_{\lambda}\|_{2^*}^{2^*} = \int_{\Bbb R^n} \left(\frac{\lambda}{{\lambda}^{2}+|x|^2}\right)^{n} dx= \int_{\Bbb R^n} \left(\frac{\lambda}{{\lambda}^{2}+|\lambda z|^2}\right)^{n} \lambda^nd z =\int_{\Bbb R^n}\frac{dz}{\left( 1+| z|^2\right)^{n}}= c_n $$
