6.21
Both numerator and denominator approach 0. So, we divide both of them by x
3 , the highest power ofx in the
denominator.
6.22
6.23
6.24
6.25
6.26
Exactly the same analysis applies as in Problem 6.23, except that, when x—» — =°, jc <0, and, therefore,
x = -Vx
2 . When we divide numerator and denominator by x and replace x by —Vx^ in the denominator, a
minus sign is introduced. Thus, the answer is the negative, -4, of the answer to Problem 6.23.
We divide the numerator and denominator by jc
3 '
2 . Note that jt
3 '
2 = V? when jc>0. So, we obtain
6.27
We divide numerator and denominator by x
2 , obtaining
38
CHAPTER 6
Find
(For a generalization, see Problem 6.43.)
Evaluate
Both numerator and denominator approach °°. So, we divide both of them by x
3 , the highest power ofx in the
dfnnminatnr.
(For a generalization, see Problem 6.44.)
Find
When/(;t) is a polynomial of degree n, it is useful to think of the degree of V/(*) as being n/2. Thus, in this
problem, the denominator has degree 1, and, therefore, in line with the procedure that has worked before, we
divide the numerator and denominator by x. Notice that, when *>0 (as it is when *-»+«), x = Vx
2 .
So. we obtain
Find
Evaluate
Evaluate
We divide numerator and denominator by x
2
. Note that
We obtain
Evaluate