1.2. Limits 5
(ii) By definition,
f Given e 0 we can find an ni(e) such that \an a\ e for all n rai(e).
Let e 0. Since n(j) j (proof by induction, if the reader demands a
proof) we have |an(j) a\ e for all j ni(e). The result follows.
(hi) Let e 0. Taking n\{e) = 1 we have
\an c\ = 0 e
for all n n\{e). The result follows.
(iv) By definition, f holds, as does
ff Given e 0 we can find an 712(e) such that \bn b\ e for all n 722(e).
Observe that
I (an + bn) - (a + 6)| = \(an - a) + (bn - b)\ \an - a\ + \bn - b\.
Thus if e 0 and /13(e) max(ni(e/2), 72 2 (e/2)) we have
I K + K) ~(a + b)\ \an -a\ + \bn - b\ e/2 + e/2 = e
for all n 723(e). The result follows.
(v) By definition, f and f| hold. Let e 0. The key observation is that
(1) \anbn - ab\ \anbn - anb\ + \anb - ab\ = \an\\bn - b\ + \b\\an - a\.
If n ni(l), then \an a\ 1 so \an\ |a| + 1 and (1) gives
(2) \anbn - ab\ (\a\ + l)|6n - 6| + |6||an - a\.
Thus
setting3
n3(e) - max (ni(l), n1(e/(2(\b\ + 1))), n2(e/(2(|a| + 1)))) we
see from (2) that
\anbn - ab\ e/2 + e/2 = e
for all n 723(e). The result follows.
(vi) By definition, f holds. Let e 0. We observe that
(3)
1
\a
aTi
\CL\\CLn
Since a ^ O w e have \a\/2 0. If 7 2 m(|a|/2) then \an - a\ |a|/2, so
\an\ |a|/2 and (3) gives
(4)
1
ari
2\a an\
\ay
3 The reader may ask why we use ni(e/(2(|6| + 1))) rather than ni(e/(2|6|)). Observe first
that we have not excluded the possibility that b = 0. More importantly, observe that all we are
required to do is to find an 713(e) that works, and it is futile to seek a 'best' ri3(e) in these or
similar circumstances.
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