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.

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