Taking a tricky limit $\lim_{p\to\infty}\int_{\Bbb R^N}\left(\frac{\left\lvert\nabla u\right\rvert}{\left\|\nabla u\right\|_p}\right)^{p-2}\cdots$ $$
\lim_{p\rightarrow \infty} \int_{\mathbb R^N} 
\left( 
\frac{\left\lvert \nabla u \right\rvert}{\left\| \nabla u \right\|_p}
\right)^{p-2} 
\frac{\nabla u}{\left\| \nabla u \right\|_p}\cdot \nabla v \,dx \label{\star}\tag{$\star$}$$
where $\,\left\lvert\,\cdot\,\right\rvert\,$ is the Euclidean $2$-norm and $\;\left\| \nabla u \right\|_p = \left\|\, \left\lvert \nabla u \right\rvert \, \right\|_p$.
Could anyone give me some hint for finding an explicit form for the expression involving $u$? I feel like my approach is not quite correct...
First I know the term in the limit is bounded
$$
\begin{aligned}
\eqref{\star} & \leq \int_{\mathbb R^N} \left( \frac{\left\lvert \nabla u \right\rvert}{\left\| \nabla u \right\|_p}\right)^{p-1} \left\lvert \nabla v \right\rvert \,dx 
\\
& = \left\| \nabla u \right\|_p^{1-p} \int_{\mathbb R^N} \left\lvert \nabla u \right\rvert^{p-1} \left\lvert \nabla v \right\rvert \,dx  
\\
&\leq \left\| \nabla u \right\|_p^{1-p} \Bigg(\int_{\mathbb R^N} \Big(\left\lvert \nabla u \right\rvert^{p-1}\Big)^{\frac{p}{p-1}}\,dx\Bigg)^{\frac{p-1}{p}} \Bigg(\int_{\mathbb R^N} \left\lvert \nabla v \right\rvert ^ p \,dx\Bigg)^{\frac{1}{p}} 
\\ 
& = \left\| \nabla u \right\|_p^{1-p}\left\| \nabla u \right\|_p^{p-1} \left\| \nabla v\right\|_p 
\\
& = \left\| \nabla v\right\|_p.
\end{aligned}
$$
with the additional assumption that $ \nabla v \in L^\infty\!\left(\mathbb R^N, \mathbb R^N\right)$.
Define the set 
$$
D:= \big\{\left\lvert \nabla u \right\rvert \leq \|\nabla u \|_\infty -\delta\big\},$$
the limit $\eqref{\star}$ is zero on the set $D$. To see this, we know there exists $N\in \mathbb N$ such that 
$$\left|\, \left\|\nabla u \right\|_\infty - \left\|\nabla u \right\|_p \,\right| \leq \frac{\delta}{2} \quad \forall \; p\geq N.$$
On the set $D$ we have 
$$ \left\|\nabla u \right\|_p - \left\lvert \nabla u \right\rvert  \geq \frac{\delta}{2} \quad \forall \; p\geq N$$
which means 
$$\frac{\left\lvert \nabla u \right\rvert}{\left\| \nabla u \right\|_p}  \leq 1 - \frac{\delta}{2} \quad \forall \; p\geq N$$
and the term in the limit $\left(\dfrac{\left\lvert \nabla u \right\rvert}{\left\| \nabla u \right\|_p}\right)^{p-2}$ goes to $0$ uniformly. 
Next I looked at the set $K: = \big\{\left\lvert \nabla u \right\rvert = \left\|\nabla u \right\|_\infty\big\}$, the term 
$$
\left(\frac{\left\lvert \nabla u \right\rvert}{\left\| \nabla u \right\|_p}\right)^{p-2} = \left(\frac{\left\| \nabla u \right\|_\infty}{\left\| \nabla u \right\|_p}\right)^{p-2}
$$
is in the form $``1^\infty"$, so I tried to use L'Hopital's rule. And we can calculate the limit (assuming the integrals are defined and finite, I just want to see what the limit might look like). 
$$
\begin{aligned}
 \lim_{p\rightarrow \infty} \left(\frac{\left\| \nabla u \right\|_\infty}{\left\| \nabla u \right\|_p}\right)^{p-2} 
&=  \lim_{p\rightarrow \infty} \exp\left( \left(p-2\right)\log\left(\frac{\left\| \nabla u \right\|_\infty}{\left\| \nabla u \right\|_p}\right)\right)
\\
&=\lim_{p\rightarrow \infty} \exp\left( \frac{\log\left(\frac{\left\| \nabla u \right\|_\infty}{\left\| \nabla u \right\|_p}\right)}{\frac{1}{p-2}}\right)
\\
&=\lim_{p\rightarrow \infty} \exp\left( \frac{\frac{d}{dp} \left[-\log\left(\frac{\left\| \nabla u \right\|_p}{\left\| \nabla u \right\|_\infty}\right)\right]}{\frac{-1}{(p-2)^2}}\right)
\\
&=\lim_{p\rightarrow \infty} \exp 
\left( 
  \dfrac{
    \left( 
       \dfrac{\left\| \nabla u \right\|_\infty}{\left\| \nabla u  \right\|_p}
    \right)
    \dfrac{
      \frac{d}{dp} 
      \big( \left\| \nabla u \right\|_p \big)
      }{ \left\|\nabla u \right\|_\infty} }
  { \dfrac{1}{\left(p-2\right)^2 }  }
\right)
\end{aligned}
$$
where
$$ 
\begin{aligned}
\frac{d}{dp}\Big[\left\| \nabla u \right\|_p \Big]  
&= \frac{d}{dp}\left[\left(\int_{\mathbb R^N} \left\lvert\nabla u \right\rvert^p \,dx \right)^{1/p} \right] 
\\
&=\frac{d}{dp}\left[\exp\left(\frac{1}{p} \log\left(\int_{\mathbb R^N} \left\lvert\nabla u \right\rvert^p \,dx \right)\right)\right] 
\\
&=\|\nabla u \|_p \left\{\frac{-1}{p^2}\log\left(\int_{\mathbb R^N} \left\lvert\nabla u \right\rvert^p \,dx \right) 
+ \dfrac{1}{p} \dfrac{1}{\int_{\mathbb R^N} \left\lvert\nabla u \right\rvert^p \,dx } \int_{\mathbb R^N} \left\lvert \nabla u \right\rvert^p \log\big(\left\lvert\nabla u \right\rvert\big) \,dx  \right\}
\end{aligned}
$$
And I am a little bit stuck.
Thank you very much!
 A: Thanks for the help from user zhw.
I claim that 
$$\lim_{p\rightarrow \infty} \int_{\mathbb R^N}\Bigg\{\left(\frac{|\nabla u|}{\|\nabla u\|_p}\right)^{p-2}\frac{\nabla u}{\|\nabla u\|_p}\Bigg\} \cdot \nabla v dx = \int_K  \frac{1}{m(K)}\frac{\nabla u}{\|\nabla u\|_\infty} \cdot \nabla v dx$$
where $K:= \{|\nabla u| = \|\nabla u \|_\infty\}$. We assume $m(K)>0$ or else the limit is $0$. 
On subset $\{\mathbb R^N - K\}$,
The term $\left(\frac{|\nabla u |}{\|\nabla u\|_p}\right)^{p-2}$ goes to $0$ pointwise a.e. on $\mathbb R^N - K$, by Lebesgue dominated convergence theorem, we have 
$$\lim_{p\to\infty} \int_{\mathbb R^N - K}  \left(\frac{|\nabla u |}{\|\nabla u\|_p}\right)^{p-2} \frac{\nabla u}{\|\nabla u\|_p} \cdot \nabla v dx = 0$$
Next on the set $K: = \{|\nabla u| = \|\nabla u \|_\infty\}$ I claim that the term
$$\lim_{p\to\infty} \left(\frac{|\nabla u |}{\|\nabla u\|_p}\right)^{p-2} = \lim_{p\to\infty} \left(\frac{\|\nabla u \|_\infty}{\|\nabla u\|_p}\right)^{p-2} = \frac{1}{m(K)}.$$ 
We can rewrite $\left(\frac{\|\nabla u \|_\infty}{\|\nabla u\|_p}\right)^{p-2}$ as
\begin{align*}
&\left(\frac{\|\nabla u \|_\infty}{\|\nabla u\|_p}\right)^{p-2} \\
=&\left(\frac{\|\nabla u \|_\infty}{\|\nabla u\|_p}\right)^{p}\frac{\|\nabla u\|_p^2}{\|\nabla u\|_\infty^2}\\
=& \frac{\|\nabla u\|_\infty^p}{\|\nabla u\|_\infty^p m(K) + \|\nabla u\|_\infty^p \int_{\mathbb R^N - K} \left(\frac{|\nabla u|}{\|\nabla u\|_\infty}\right)^p dx} \frac{\|\nabla u\|_p^2}{\|\nabla u\|_\infty^2}\\
=& \frac{1}{ m(K) + \int_{\mathbb R^N - K} \left(\frac{|\nabla u|}{\|\nabla u\|_\infty}\right)^p dx} \frac{\|\nabla u\|_p^2}{\|\nabla u\|_\infty^2}
\end{align*}
Take the limit $p\to \infty$ and by Lebesgue dominated convergence theorem we get
$$\lim_{p\to\infty} \left(\frac{|\nabla u |}{\|\nabla u\|_p}\right)^{p-2} = \frac{1}{m(K)}$$
on the set $K$.
Thus the integral 
$$\lim_{p\to\infty} \int_K \left(\frac{|\nabla u |}{\|\nabla u\|_p}\right)^{p-2} \frac{\nabla u}{\|\nabla u\|_p} \cdot \nabla v dx =\int_K  \frac{1}{m(K)}\frac{\nabla u}{\|\nabla u\|_\infty} \cdot \nabla v dx$$
