2003 Test 1 solution

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10 Apr 2012
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MAT 137Y, 2003-2004, Solutions to Test 1
1. Evaluate the following limits.
(8%) (i) lim
t1
t31
t1.
lim
t1
t31
t1=lim
t1
(t1)(t2+t+1)
t1=lim
t1t2+t+1=3.
(8%) (ii) lim
x3
x251+x
x29.
Let Lbe the limit. Then
L=lim
x3
(x251+x)(x25+1+x)
(x29)(x25+1+x)
=lim
x3
x251x
(x29)(x25+1+x)=lim
x3
x2x6
(x29)(x25+1+x)
=lim
x3
(x+2)(x3)
(x+3)(x3)(x25+1+x)=lim
x3
(x+2)
(x+3)(x25+1+x)=5
24.
(12%) 2. Solve the inequality |x3|+x9.Express your answer in interval notation.
If x3, then x3+x9, or x6. Therefore, a solution is x[3,6].
If x<3, then 3 x+x9, or 3 9, which is true for all xwhich satisfies the assumption, so
x(,3)also satisfies the inequality. Combining solutions, we get x(,6].
3.
(5%) (a) Give the formal ε,δdefinition of the statement lim
xaf(x) = L.
For any ε>0, there exists δ>0 such that 0 <|xa|<δimplies |f(x)L|<ε.
(8%) (b) Suppose for all x,|f(x)| ≤ B, where Bis some positive number. Prove directly from the ε,δ
definition that lim
x0xf(x) = 0.
We need to show that for any ε>0 there exists δ>0 such that 0 <|x|<δimplies |xf(x)|<ε.
Given ε>0, choose δ=ε
B. Then 0 <|x|<δimplies |x|<ε
B, so |xf(x)|=|x|·|f(x)| ≤ δ·B=
ε
B·B=ε, so |xf(x)|<ε, which is what we needed to show.
(12%) (c) Prove directly from the ε,δdefinition that lim
x3
x
1+x2=3
10.
We need to show that for any ε>0 there exists δ>0 such that 0 <|x3|<δimplies |x
1+x2
3
10|<ε. Choose δ=min(1,50
11ε). Then δ1 and δ50
11ε. If δ1, then |x3|<δ=
|x3|<1=2<x<4, so |3x1|<11 and
1
10(1+x2)
<1
50. Thus,
0<|x3|<δ=
x
1+x23
10
=
10x33x2
10(1+x2)
=
(3x1)(x3)
10(1+x2)
=
1
10(1+x2)
·|3x1|·|x3|<1
50 ·11·δ11
50 ·50
11ε=ε,
so |x
1+x23
10|<ε, as required.
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