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

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

Now, I've already shown that the above implies that $|A-B| < \epsilon$, 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 $A\ne B$.
Then in what you did above let $\epsilon=|A-B|>0$.
You last line says $|A-B|<\epsilon=|A-B|$ that is a contradiction.