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University of Illinois at Urbana Champaign Fall

3 pages
University of Illinois at Urbana-Champaign Fall 2006 Math 444 Group E13 Training exercises : Correction. 5.2.7 : We more or less saw that example in class : define a function f : [0, 1] ? [0, 1] by setting, for all x ? [0, 1], f(x) = { ?1 if x ? Q 1 else . Pick a number x ? [0, 1], and recall that there exist a sequence (qn) of rational numbers in [0, 1 and a sequence (?n) of irrational numbers in [0, 1] such that lim(qn) = lim(?n) = x. One also has f(qn) = 0 for all n ? N, and f(?n) = 1 for all n inN ; this proves that f doesn't have a limit at x, so it cannot be continuous at that point. Thus f is discontinuous at every point of [0, 1], yet |f | is constant (equal to 1), so it is a continuous function. 5.2.8 : Recall again that any real number is the limit of a sequence of rational numbers ; pick x ? R, and a sequence of rational numbers (qn) that converges to x. Then one has lim f(qn) = f(x) since f is continuous at x, and for the same reason lim g(qn) = g(x).

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1 .
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f(q ) = 0 n∈N f(α ) = 1 n inN fn n
x f [0,1] |f|
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g(x) = 0 x∈R g(0)g(x) = g(x) x inR
g(0) = 1
g 0 x∈R (x ) xn
limg(x ) = g(x) g 0n
0
(y ) = (x −x)n n
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limg(y ) = g(0) = 1 g 0 limg(x ) = g(x)n n
(x ) g x x ∈ R gn
R
g g(0) = 1 g
h(x) = ln(g(x)) h(x +y) = h(x) +h(y) x ∈ R
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−210
[189/128,190/128]
−210 1.48

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