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For american put option I have to prove that:

1) As $D$ tends to $\infty$, $a_n$ tends to $-r/D$ so that $S^*$ tends to $0$.

2) As $D$ tends to $-\infty$, $a_n$ tends to $2D/ \sigma^2$ so that $S^*$ tends to $E$.

I have simplified the question posted here american put option

to the point where $a_n =\frac1{\sigma^2}\lim_{D\to\infty}[-\sqrt{D^2 + 2 r \sigma^2} + D]$.

I think to prove 1) and 2) above I have to do a binomial expansion on the square root keeping only the first two terms in the expansion

I am not sure how do I go about this. Any help would be greatly appreciated.

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marked as duplicate by Did, Git Gud, Davide Giraudo, Martin, user1551 Apr 14 '13 at 12:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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This seems like an obvious auto-duplicate and as such, should be closed. I posted a full answer on the other page, using only the conjugate quantity. –  Did Apr 14 '13 at 11:18

1 Answer 1

up vote 1 down vote accepted

Note that $\sqrt{1+x}=1+\frac12x+O(x^2)$ as $x\to 0$, hence $$\begin{align}D-\sqrt{D^2+2r\sigma^2} &= D\cdot\left(1-\sqrt{1-\frac{2r\sigma^2}{D^2}}\right)\\&=D\cdot\left(\frac{2r\sigma^2}{D^2}+O(D^{-4})\right)\\ &=\frac{2r\sigma^2}D+O(D^{-3}) \end{align}$$ as $D\to+\infty$. By the same reasoning, $$D-\sqrt{D^2+2r\sigma^2}=2D+O(D^{-1})$$ as $D\to-\infty$.

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Hi Hagen, Please can you elaborate on the second one? By the same reasoning, $$D-\sqrt{D^2+2r\sigma^2}=2D+O(D^{-1})$$ as $D\to-\infty$. –  sleepybob Apr 20 '13 at 15:40

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