MAT 137Y, 2003-2004, Solutions to Test 1
1. Evaluate the following limits.
(8%) (i) lim
t→1
t3−1
t−1.
lim
t→1
t3−1
t−1=lim
t→1
(t−1)(t2+t+1)
t−1=lim
t→1t2+t+1=3.
(8%) (ii) lim
x→3
√x2−5−√1+x
x2−9.
Let Lbe the limit. Then
L=lim
x→3
(√x2−5−√1+x)(√x2−5+√1+x)
(x2−9)(√x2−5+√1+x)
=lim
x→3
x2−5−1−x
(x2−9)(√x2−5+√1+x)=lim
x→3
x2−x−6
(x2−9)(√x2−5+√1+x)
=lim
x→3
(x+2)(x−3)
(x+3)(x−3)(√x2−5+√1+x)=lim
x→3
(x+2)
(x+3)(√x2−5+√1+x)=5
24.
(12%) 2. Solve the inequality |x−3|+x≤9.Express your answer in interval notation.
If x≥3, then x−3+x≤9, or x≤6. Therefore, a solution is x∈[3,6].
If x<3, then 3 −x+x≤9, 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
x→af(x) = L.
For any ε>0, there exists δ>0 such that 0 <|x−a|<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
x→0xf(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
x→3
x
1+x2=3
10.
We need to show that for any ε>0 there exists δ>0 such that 0 <|x−3|<δimplies |x
1+x2−
3
10|<ε. Choose δ=min(1,50
11ε). Then δ≤1 and δ≤50
11ε. If δ≤1, then |x−3|<δ=⇒
|x−3|<1=⇒2<x<4, so |3x−1|<11 and
1
10(1+x2)
<1
50. Thus,
0<|x−3|<δ=⇒
x
1+x2−3
10
=
10x−3−3x2
10(1+x2)
=
(3x−1)(x−3)
10(1+x2)
=
1
10(1+x2)
·|3x−1|·|x−3|<1
50 ·11·δ≤11
50 ·50
11ε=ε,
so |x
1+x2−3
10|<ε, as required.
1