ANSWERS AND SOLUTIONS TO SELECTED EXERCISES^367- (b) Proof Assume C is a set satisfying C E A, for each k = 1,2,3,.... To
prove C c ng, A,, let x E C and let j be an arbitrary positive integer; we
must prove x E Aj. Since C c A, for each k = 1,2,3,... , then, in particular,
C E Aj. Since x E C and C c Aj, we have x E Aj. Since j was arbitrarily
chosen, we have x E n~= A,, as desired. 0
Article 4.3
- (a) (b) (iii) lirn,,, f(x) = = 4. Type I (i.e., continuous) at a = 3.
(uiii) lirn,,, f (x) = lirn,,, ((3 - x)/3x)(l/(x - 3)) = lirn,,, (-$x) = -$. Type
I1 (i.e., not continuous) at a = 3 since f is not defined at 3. - (a) lirn,, , f (x) does not exist since lirn,, , - f (x) = 9 # 1 = lim,, , + f (x).
(b) lirn,,, f(x) = lirn,,, (2x - 5) = 1. - (a) (iii) Let E be any specific positive real number less than or equal to B (i.e.,
B is the largest value of E that could be used), say, E = B/2. Let 6 be an
arbitrary positive real number. Then, for any nonzero x between -6 and
6,wehaveeitherO<x<6and f(x)=5x+B>B>B/2,orelse -6<x<O
and f(x) = 2x - B < - B < - B/2. In either case f(x) is not within a distance
of E = B/2 of L = 0. Surely then, corresponding to the given (specific) E and
an arbitrarily chosen positive 6, there can always be found x such that
0 -c 1x1 c 6, whereas 1 f(x)l 2 E. - (d) The graph of the tangent function tells us that lim,,(,,,,- = m, whereas
limx,(r,2)+ = - m. - (a) Let E = 3 and let 6 > 0 be given. Corresponding to this arbitrarily chosen
6, there clearly exists a positive integer n such that both 2/(4n + 1)n and
2/(4n + 3)n are less than 6. If x = 2/(4n + 1)n, then l/x = (4n + 1)71/2 so that
sin (l/x) = sin [(4n + 1)71/2] = 1. If x = 2/(4n + 3)n, then l/x = (4n + 3)42 so
that sin (llx) = sin [(4n + 3)n/2] = - 1. We conclude that, in any
neighborhood of a = 0 (no matter how narrow), there exist values of x for
which sin (l/x) = 1 and sin (llx) = - 1. Thus it is impossible, given an E
neighborhood of L = 0 with E < 1, to find a 6 neighborhood about a = 0
such that all positive values of x within that neighborhood have their
corresponding f(x) within E of L = 0. - (b) (ii) lirn,,, (x + 6)/(x2 - 10) = lirn,,, ((llt) + 6)/((l/t2) - 10) =
lirn,,, ((6t + l)/t)(t2/(1 - lot2)) = lim,,, (6t2 + t)/(l - lot2) = 011 = 0.
(iii) lirn,,, (x2 - 2x + 3)/(2x2 + 5x - 3) = lirn,,, [(I - (3/x) +
(3/x2)]/[(2 + (SIX) - (3/x2)] = (1 - 0 + 0)/(2 + 0 - 0) = 4.
(d) (i) lirn,,, f(x) = m if and only if, corresponding to any M > 0, there
exists 6 > 0 such that, whenever 0 < Ix - a1 < 6, then f(x) 2 M.
Article 5.1
- (a) (i) Proof Let x be an arbitrary real number. Then, cos4 x - sin4 x =
(cos2 x - sin2 x)(cos2 x + sin2 x) = (cos2 x - sin2 x)(l) = cos2 x - sin2 x =
cos 2x. 0