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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.

Stella Biderman
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Dora Sengupta
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  • Stating the obvious perhaps, but the problem is reduced to showing that $\lim_{n\rightarrow \infty}[(1-u)^{\frac1n}+v^{\frac1n}-1]^n=v-uv$. Perhaps the Multinomial Theorem will help with the power of sums? – MathematicianByMistake Apr 18 '16 at 20:25

1 Answers1

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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,

$$\lim_{n\rightarrow \infty} n\cdot \ln[(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]=\lim_{n\rightarrow \infty}\dfrac{n}{\frac{1}{\ln[(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]}}=\lim_{n \rightarrow \infty} \frac{d}{dn}\dfrac{n}{\frac{1}{\ln[(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]}}$$ $$=\ln(1-u)+\ln(v)=ln(Y) $$

$Y=(1-u)\cdot v=v-uv$

Beth
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  • Could you please clarify a few points on this approach? How does $\lim_{n\rightarrow \infty} n\cdot \ln[(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]=0\cdot\infty$? How do you differentiate $\lim_{n \rightarrow \infty} \frac{d}{dn}\dfrac{n}{\frac{1}{\ln[(1-u)^{\frac{1}{n}}+v^{\frac{1}{n}}-1]}}$ And where do the constaints imposed play a part? Thanks in advance. – MathematicianByMistake Apr 18 '16 at 22:01