Prove:

lim $\displaystyle a_{n}$ > M --> $\displaystyle a_{n}$ > M for n>>1 (large n).

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- Sep 17th 2009, 12:17 PMcgiulzlimit > M, then An> M?
Prove:

lim $\displaystyle a_{n}$ > M --> $\displaystyle a_{n}$ > M for n>>1 (large n). - Sep 17th 2009, 02:27 PMxalk
- Sep 17th 2009, 04:56 PMcgiulz
for large n.

- Sep 17th 2009, 07:33 PMxalk
Then we have:

Since $\displaystyle \lim_{n\rightarrow\infty} a_{n} =l$>M THAT implies that for all ε>0 and hence for ε= l-M>0 ,there exists an N such that :

for all n , $\displaystyle n\geq N$ then $\displaystyle |a_{n}-l|< l-M$

or $\displaystyle M-l< a_{n}-l< l-M$.

Thus $\displaystyle a_{n}> M$ for large n biger or equal to N