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University of Illinois at Urbana Champaign Fall

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Niveau: Supérieur, Doctorat, Bac+8
University of Illinois at Urbana-Champaign Fall 2006 Math 444 Group E13 Graded Homework X. Due Friday, November 17. 1.(a) Define a function f on R by setting f(x) = { f(x) = x sin( 1x ) if x 6= 0 0 else . Is this function continuous on R ? (you may use without proof the fact the the function x 7? sin(x) is continuous). (b) Let g : R? R be defined by g(x) = { x if x ? Q 1? x else . At which points in R is g continuous ? Correction. (a) Since we are told that x 7? sin(x) is continuous on R, we see immediately that f is continuous on (?∞, 0)?(0,+∞) : indeed, on these two intervals f is a composition of continuous functions. Thus, our only problem is at x = 0 ; we need to see whether f has a limit at 0, and whether that limit is equal to f(0) = 0. As usual with sin, the important thing to remember is that | sin(x)| ≤ 1 for all x ? R ; thus, for all x 6= 0 we see that |f(x)| ≤ |x|.

  • urbana-champaign fall

  • must then

  • graded homework

  • squeeze theorem

  • continuous

  • functions

  • there exists


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x1. f R f(x) =
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R x7→ sin(x)(
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x = 0 f 0 f(0) = 0
sin |sin(x)|≤ 1 x∈R x = 0
|f(x)|≤|x| lim f(x) = 0
x→0
lim f(x) = f(0) f 0 f
x→0
x∈R (q ) (α )n n
x = lim(q ) = lim(α ) g x g(x) = lim(f(q )) = lim(q ) = xn n n n
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[a,b] x∈ [a,b] ε > 0 y ∈ [a,b]
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δ = ε f
f x x∈ I f I
g(x) = f(x)−x g(a) = f(a)−a≥ 0 a≤ g(x)≤ b x∈ [a,b]
g(b) = f(b)−b≤ 0 g(a) = 0 g(b) = 0 g a
b x∈ [a,b] g(x) = 0
f [a,b]
x ∈ [a,b] f(c) = x g(x) = 0 f(x) = x
0x = x ∈ [a,b] f(x) = x
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1 1
n∈N x∈ [0,1− ] f(x) = f(x+ )
n n
f((k +1)/n)−f(k/n) k = 0,...,n−1
n≥ 2 n = 1 f(0) = f(1)
1 1
f((k +1)/n) = f(k/n) k = 0,1,...,n−1 x∈ [0,1− ] f(x) = f(x+ )
n n
k/n f((k +1)/n) = f(k/n) k = 1,...,n−1
k∈ 0,...n−2 f((k +1))/n−f(k/n) f((k +2)/n)−f(k +1/n)
f(1/n) > f(0) f(2/n) > f(1/n)
f(3/n) > f(2/n) f(1) = f(n/n) > f((n− 1)/n) > ...f(0)
f(1) = f(0) f(1/n) < f(0)
k f((k +1))/n−f(k/n) f((k +2)/n)−f(k +1/n)
g(x) = f(x+1/n)−f(x) g [k/n,(k +1)/n]
k = 0,...,n− 2 g 0 0 g
x f(x) = f(x + 1/n) f
x∈ [0,1−1/n] f(x) = f(x+1/n)
4. f [a,b] f(x) > 0
x∈ [a,b] m > 0 f(x)≥ m x∈ [a,b]
f,g [0,1]→ [0,1] f(x) < g(x) x∈ [0,1]
m > 0 f(x)+m < g(x) x∈ [0,1]
M > 1 Mf(x) < g(x) x∈ [0,1]
f [a,b]
[a,b] c ∈ [a,b] f(c) ≤ f(x) x ∈ [a,b]
m = f(c) m > 0 f f(x)≥ m x∈ [a,b]
0g−f [0,1] m > 0
0m0g(x)−f(x)≥ m x∈ [0,1] m = m > 0 g(x)−f(x) > m
2
0x∈ [0,1] f(x)+m < g(x) x∈ [0,1] m
2
0 ≤ f(x) < g(x) x ∈ [0,1] g
f 0h = [0,1] h(x) < 1 x ∈ [0,1] m > 0
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0m 0[0,1] m = h(x) < 1−m
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100 0m 0 < m < x∈ [0,1] h 1−m > 0 g(x) > 0
2
1 1
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0 01−m 1−m
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(q ) lim(q ) = x fn n
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x∈R
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