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Unformatted text preview: 15. 16. 17. 18. 19. 20. 0 < x — 2 < 6, then if x is rational, f(x  16 = 8X  16] = 8[x  2 < 86 S 6. If x is irrational, f(x)  16’ = 2x2 + 8  16 2 2):2 — 41: 21(x— 2)(x+ 2) 510x 2 < 1065 .5.
Note that [x + 2 5 5 since 6 was chosen to be less than or equal 1.
The key to this problem is to note the similarity between this theorem and the theorem that states that Cauchy
sequences are convergent. In fact, let’s use that theorem. Let {an}'::___1 be any sequence in D\{x0} converging to x0. Choose 6 > 0. There is a neighborhood Q of XO
such that xgy E Q n D, x % x0 and y % x0 implies that
f(x)  f(y& < 6. There is N such that n 2 N implies
that an E . Thus if m,n 2 N, an, an E Q n D, an % x0, am % x0, then f(an)  f(am){ < e. We may conclude that the sequence {f(an)}§=1 is Cauchy, hence convergent.
Theorem 2.1 then guarantees that f has a limit at no. 3 2 2
Since dom f = (0,1), f(x) = x + 6x + x ; x ; 6? + 1
x2 — 6x 6
since 0 E dom f) for all x 6 0,1). But then we apply heorem 2.4 to the numerator and enominator to conclude
1 that f has a limit at 0 and that limit is  6. iff x0 is not an integer, the analysis in The function has a limit at x0 is not an odd
integer. If x0 Exercise 13 will suffice. If x0 is an odd integer, use a sequence such as that in Example 2.6 to show that f does
not have a limit at x0. If x0 = 2n, then for
2n  1 < x < 2n + 1, show that {(x) = x  2n.
For x=f=0,g(x):1—1::_’_1=___x—_=__1_
x(41 + x + 1) 41 + x + 1
The limit is For x ,6 o, {(x) : s—_'1— , the limit is  g
49 — x + 3
Apply Theorem 2.1 and Theorem 1.12. 25 El. 23. in essence, we want to show that in a neighborhood of g is bounded away from zero. The theorem should be
something along the following lines: X0, Suppose point of D and g(x)
g has a limit as x0 Theorem: g:D a R with x0 an accumulation % 0 for all x E D. Assume that
and lim g(x) % O.
XDi 0
Then there is a positive real number M and a neighborhood
Q of x0 such that x E Q n D, x $ x0 implies that
lg(x)l 21“
Proof: Let L = lim g(x) and e = '% There is a
xaxo ' neighorhood Q of XO such that x E Q n D, x # x0, implies that g(x) — L[ < e. Let M = 1%i. Then if x E Q n D and x # X0, ]g(x) = g(x)  L + L} 2le lamLl ZlLlLIfL=J—I§i=m. (a) If f(x) : i + 6 and g(x) =  i + 3, X $ 0, both f and g fail to have a limit at zero, whereas f + g does have a limit at zero. (b) Define f(x) = i for —1 S x < O and f(x) = 0 for 0 < x S 1. Define g(x) 2 0 for ~1 S X S 0 and g(x) = % for O < x S 1. Both fail to have a limit at zero, yet
(fg)(x) : 0 for all —1 S x < 1. (c) Define f(x) = i for O < x S 1 and g(x) = l? for
X 0 < x S 1. Both fail to have a limit at zero but g has limit zero at zero. Let f:[a,5 A R be decreasing. Let C(x) = inf f(y):y < x} and L(x) = sup {f(y):x < y for each x E (0,5). Then f has a limit at x0 E (a,ﬂ iff U(x0) = L(x0) and in this case, U(x0) = L(x0) = lim f(x)
X—ix 0 : f(x0). Proof: Suppose f has a limit at x0, call it A. Choose 26 24. 25. e > 0. There is 6 > 0 such that if 0 < {x  x0 < 6
with x 6 [0,6], then f(x)  Al < 6. Since x0 E (0,6)
there are x,y 6 [6,6] such that x0 * 6 < x < x0 < y < x0 + 6 so A  e < f(y) S L(x0) S f(x0) S U(x0) S f(x) < A+ 6. Since € > 0 was arbitrary, we must
conclude that A = U(x0) = L(x0) = f(x0). Suppose now that U(x0) = L(x0). we noted that L(x0) S_f(x0) S U(x0), hence L(x0) = f(x0) = U(x0).
Choose 6 > 0. There are numbers y1,y2 such that a S yl < x0< y2S ﬂ and such that L(x0)  6 < f(y2) and
f(y1) < U(x0) + e. Let 6 = min {x0  y1,y2 ~ x0}. Now if
0 < x  x0] < 6, then yl < x < y2 and f(x0)  e = L(x0)  e < f(y2) 3 f(x) s f(y1) < U(x0) + e = f(x0) + 6. Thus f has a limit at x0 and
that limit is f(x0). Let’s do the case for f increasing. The case for f
decreasing is similar. Let B = sup {f(x):a < x < b}. Choose 6 > 0. B  6 is
not the least upper bound of the set {f(x):a S x < b},
hence there is x1 6 [a,b) such that b  6 < f(x1). Since f is increasing, x1 < x < b implies that
B  6 < f(x1) < f(x) S B < B + 6. Thus f has limit B at b. Likewise define A = inf {f(x):a < x < b}.
the limit of f at a. The details are similar. This is Since g is increasing, Exercise 24 guarantees that g
has a limit at a and at b, so we will examine only
a < x0 < b. Define L(x0) and U(x0) (for the function g) as in Lemma 2.7. We will show that L(x0) = g(x0) = U(x0). The roof will be by contradiction. Suppose
L(x0§ < g(x0). Since f(t) S g(t) for all a S t < x0, f(t) S L(x0) for all a S t < x0. Therefore,
lim {(t) = f(x0) S L(x0). But this contradicts
t—ixO
g(x0) =
g(x0) < U(xo).
Now for each y S g(¥)
such that x0 < z < y sup {f(t): A S t S x0} > L(x0). Suppose now that
Chooose s such that g(x0) < s < U(x0).
such that x0 < y < b, g(x0) < s < U(x0) and so by the definition of g(y), there is z
and s < f(z). So we have 27 26. 27. f(x0) S g(x0) < s < f(z) contrary to lim f(t) = f(x0). taxo The function f(x = ex is an example to keep in mind
while attaching t is problem. Suppose f has a limit at zero, call it L. Note
that f(x) = féx + 0) = f(x)f(0) for all x E R, hence
f§0) = 1 or (x) = 0 for all x E R. If f(x) : 0 for
a 1 x E R, then the problem is finished. Suppose f(x) % 0 for some x E R. It follows that
f(0) = 1. Let L be the limit of f at 0. Note that f(2x) = [f(x)]2 and by induction, f(nx) = [f(x)]n. The
sequence {ﬁ}:=1 converges to zero, hence the sequence 1
{f(§)}::1 converges to L. But f(§) = [f(x)]rl and by
l m
results from Chapter 1,{[f(x)]n} converges to 1, hence
n=1 L = 1. We now know that lim f(x) = 1.
Xﬁo f(x) = f(x  x0)f(x0) and the function h(x) = f(x  x0) has a limit at x0, hence so does f. If x0 # 0, note Since x0 is an accumulation point of D n E, it is also
an accumulation point of D as well as an accumulation
point of E. Let {xn}::1 be any sequence in There is N such that
Thus for n 2 N, (D n E)\{x0} converging to x0. n > N implies that [xn  x0] < e. f(xn) = g(xn). Therefore, the sequence {f(xn)}§=1
converges iff {g(xn)}:=1 converges. Hence f has a
limit at x0 iff g has a limit at x0. This is a rather subtle point, but perhaps yoa might
want to discuss it with your students. It says that the
property of having a limit is a local condition, i.e., the
behavior near x0 is all that is important. PROJECT 2.1 1. This is almost trivial proof by induction, but perhaps it
is worthwhile to insist that your students do it. 28 “ "’Wﬁ'tmrw mmv‘wrﬁ: "hm ,. .i ...
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 Spring '07
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