1. ## Sequence Limit Question

First it states to give an alternate proof of a theorem that showed that a sequence's limit is unique.

Choose $\displaystyle \epsilon > 0$. There is an $\displaystyle N_1$ such that for $\displaystyle n \geq N_1$, $\displaystyle |a_n - A| < \frac{\epsilon}{2}$ and there is an $\displaystyle N_2$ such that for $\displaystyle n \geq N_2$, $\displaystyle |a_n - B| < \frac{\epsilon}{2}$. Use the triangle inequality to show that this implies that $\displaystyle |A-B| < \epsilon$. Argue that $\displaystyle A = B$.

Now, I've already shown that the above implies that $\displaystyle |A-B| < \epsilon$, but I'm having trouble seeing how I can argue that the limits are equal (even though I know they are).

2. ## Re: Sequence Limit Question

Originally Posted by Aryth
First it states to give an alternate proof of a theorem that showed that a sequence's limit is unique.

Choose $\displaystyle \epsilon > 0$. There is an $\displaystyle N_1$ such that for $\displaystyle n \geq N_1$, $\displaystyle |a_n - A| < \frac{\epsilon}{2}$ and there is an $\displaystyle N_2$ such that for $\displaystyle n \geq N_2$, $\displaystyle |a_n - B| < \frac{\epsilon}{2}$. Use the triangle inequality to show that this implies that $\displaystyle |A-B| < \epsilon$. Argue that $\displaystyle A = B$.

Now, I've already shown that the above implies that $\displaystyle |A-B| < \epsilon$, but I'm having trouble seeing how I can argue that the limits are equal (even though I know they are).
Start by supposing that $\displaystyle A\ne B$.
Then in what you did above let $\displaystyle \epsilon=|A-B|>0$.
You last line says $\displaystyle |A-B|<\epsilon=|A-B|$ that is a contradiction.