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.