082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

6.2 Sequences 207
can sensibly letk → ∞. Dividing both numerator and denominator byk, we write
k −1 1− (1/k)
=
k +1 1+ (1/k)
Then, ask → ∞, 1/k → 0 so that
( ) 1− (1/k)
lim
k→∞
1+ (1/k)
= lim k→∞
(1 − (1/k))
lim k→∞
(1 + (1/k))
byrule (4)
= 1 1
= 1
Example6.10 Given a sequencewith generalterm
x[k] = 3k2 −5k+6
k 2 +2k+1
find lim k→∞
x[k].
Solution Dividing the numerator and denominator byk 2 introduces terms which tend to zero as
k → ∞, that is
3k 2 −5k+6
k 2 +2k+1 = 3 − (5/k) + (6/k2 )
1 + (2/k) + (1/k 2 )
Then ask → ∞, wefind
lim
k→∞ x[k] = 3 1 = 3
k
Example6.11 2
Examine the behaviour of
3k+1 ask→∞.
Solution k 2
3k+1 = k
3+ (1/k)
As k → ∞, 1/k → 0 so that the denominator approaches 3. On the other hand, as
k → ∞the numerator tends toinfinity so thatthissequence diverges toinfinity.
EXERCISES6.2
1 Graphthe sequences given by
(a)x[k]=k,k=0,1,2,3,...
{ 3 k = 2
(b)x[k] = k=0,1,2,3,...
0 otherwise
(c)x[k]=e −k ,k=0,1,2,3,...
2 Thesequencex[k] is obtainedbysampling
f (t) = cos(t +2),t ∈ R.Thesampling begins at
t = 0 andthereafter att = 1,2,3, . . ..Writedown
the firstsixterms ofthe sequence.
3 A sequence,x[k], is defined by
x[k] = k2
2 +k,k=0,1,2,3,...
State the firstfiveterms ofthe sequence.
4 Writedownthefirstfiveterms,andplotgraphs,ofthe
sequences given recursively by
(a) x[k] =
x[k −1]
, x[0]=1
2