epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Solutions for the Exercises at the End of the Sections and the Review Exercises 171
and
—e — -<L<s — ~.
Because s can be any positive number, they will have to hold true
even when s = 1/5. In this case, the first set of inequalities will yield
0 < L and the second L< 0. Clearly this is impossible. Therefore,
the limit of the sequence does not exist.