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Oct13-09, 12:48 PM
HW Helper
PF Gold
LCKurtz's Avatar
P: 7,659
You have the right idea, but it isn't written well. You need to use your observations to write your argument in reverse. Something like this:

Given can [itex]\rightarrow[/itex] L we will show that an [itex]\rightarrow[/itex] L/|c|, which is a contradiction.

Suppose[itex]\ \epsilon > 0[/itex]. Then there is N > 0 such that:

| can - L| < |c|[itex]\epsilon [/itex]

for all n > N. Therefore

|can - L| = |c(an - L/c| = |c||(an - L/c| < |c|[itex]\epsilon[/itex]

which gives, upon dividing that last inequality by |c|, | an - L/c| < [itex]\epsilon[/itex]
for all n > N.