Proof of a limit problem Let $C_{n}(u,v)=v-[max\{(1-u)^{1/n}+v^{1/n}-1, 0\}]^{n}$ under the constraints that $0 \leq u \leq 1, 0 \leq v \leq 1$.
Prove that, $$\lim_{n\to\infty} C_{n}(u,v)=uv$$
Thanks in advance.
 A: As mentioned in the comment, it becomes evident to show that 
$$\lim_{n\rightarrow \infty} [(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]^n=v-uv$$
under the constraints.
Let $Y=\lim_{n\rightarrow \infty} [(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]^n$
So, $\ln(Y)= \ln\left[\lim_{n\rightarrow \infty} [(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]^n\right]=\lim_{n\rightarrow \infty} \ln\left([(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]^n\right)$
$\quad\quad\quad\quad=\lim_{n\rightarrow \infty} n\cdot \ln[(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]$
Note that $\lim_{n\rightarrow\infty} n^{\frac{1}{n}}=1$
It follow that $\lim_{n\rightarrow \infty} n\cdot \ln[(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]=0\cdot\infty$
By applying L'Hopital's Rule on the second expression below,
$$\lim_{n\rightarrow \infty} n\cdot \ln[(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]=\lim_{n\rightarrow \infty}\dfrac{n}{\frac{1}{\ln[(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]}}=\lim_{n \rightarrow \infty} \frac{d}{dn}\dfrac{n}{\frac{1}{\ln[(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]}}$$
$$=\ln(1-u)+\ln(v)=ln(Y)  $$
$Y=(1-u)\cdot v=v-uv$
