Calculus: Analytic Geometry and Calculus, with Vectors

3.3 Unilateral limits and asymptotes 141

17 We can feel sure that if IxI < 1, then x" is near 0 whenever n is large.
How can we prove it? Solution: Let e > 0. Suppose first that x = 0. Then
ix"I < e when n > 1. Suppose finally that 0 < jxj < 1. Let y = 1/IxI so that
y > 1. Then the preceding problem shows that

Jim y" = co.

If we choose an index N such that y" > 1/e when n > N, then

jx"i=y<e

when n > N. Therefore,

18 Prove that

lim x" = 0

n-.0

lim 1 + I +23+. T2 +fl = 1

,,

(IxI < 1).

Jim [1 + x + x2 + ... + x"] =1 1

x (IxI < i).

Hint: Long division (or factoring) shows that

1 - x"+1

l+x+x2-}- 1-x

and we may use the fact that lim x" = 0 when IxI < I.

19 Once again, let the "bracket symbol" [x] denote the "greatest integer

in x," that is, the greatest integer less than or equal to x, so that [8] = 8and

[15.359] = 15. Show that, for each integer n,

lim [x]=n, lim [x]=n-1.

z- n+ X--+n-

20 Prove that if g is a function and .4 and B are numbers such that jg(x)j S 11

whenever x > B, then

limg(x)=0

and that, if L is a number, then

21 Prove that

lim (L +g(x)1= L.

X-.( x JJ

1Jm[x]= 1.

x X

Hint: Let 0(x), read "theta of x," denote the "fractional part" of x sothat

8(x) = x - [x] and [x] = x - 8(x).

22 Sketch a graph of the function h for which h(x) = [x]/x when x >=1,

and observe that h(x) really is near 1 whenever x is large.