When trying to prove thar the limit of a seq is unique, i saw in my notes tt for episilion >0, la-bl>2episilion assuming that a and b r the limits of the seq.

Then from here, why does it mean that la-bl =0?

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- Oct 21st 2010, 01:02 AMalexandrabel90Epsilion
When trying to prove thar the limit of a seq is unique, i saw in my notes tt for episilion >0, la-bl>2episilion assuming that a and b r the limits of the seq.

Then from here, why does it mean that la-bl =0? - Oct 22nd 2010, 06:23 PMTinyboss
If you mean $\displaystyle |a-b|<2\varepsilon$ then we get equality because $\displaystyle \varepsilon$ can be arbitrarily small. If a and be were not equal you could choose $\displaystyle \varepsilon$ small enough to break the inequality. Note that any constant multiple of $\displaystyle \varepsilon$ would work just as well.