Download M361K Homework #7. Problem 1.) Prove the following limits a.) lim

M361K Homework #7.
Problem 1.) Prove the following limits
a.)
lim x2 − 2x + 2 = 2
x→2
Solution: (Given > 0 we need to find δ > 0 so that |x2 − 2x + 2 − 2| < when
0 < |x − 2| < δ.) Given any > 0 choose δ = min{1, /3}. Then, if 0 < |x − 2| < δ, it
must be that |x − 2| < 1 and therefore |x| < 3. Now,
|f (x) − 2| = |x2 − 2x + 2 − 2| = |x||x − 2| < 3|x − 2|
so if |x − 2| < δ, then |x − 2| <
3
and |f (x) − 2| < .
b.)
lim
x→0
1
1
=
2+x
2
1
Solution: (Given > 0 we need to find δ > 0 so that | 2+x
− 12 | < when
0 < |x − 0| < δ.) Given any > 0, choose δ = min{1, }. Then, if 0 < |x| < δ, it must
1
be that |x| < 1 so 1 < x + 2 < 3 and |x+2|
< 1. On the other hand,
1
1
|x|
|x|
1
− |=
<
|f (x) − | = |
2
x+2 2
2|x + 2|
2
so if |x| < δ, then |x| < 2 so |f (x) − 21 | < .
c.)
lim f (x) = 0
x→2
where f (x) = (x − 2)2 if x ≤ 2 and f (x) = (x − 2)3 if x > 2.
Solution: (Given > 0 we need to find δ > 0 so that |f (x)| < when
0 < |x − 2| < δ. We treat
√ the
√ function different from the right or the left.)
√ Given
> 0 choose δ = min{ , 3 }. If x > 2 and |x − 2| < δ, then |x − 2| < 3 and
√
|f (x)| = |x|3 < . On the other hand, if x < 2 and |x − 2| < δ, then |x − 2| < and
|f (x)| = |x|2 < . Either way, |f (x)| < .
To solve the next problem we use the following negation of the definition of the limit.
Definition 0.1. We say a function f : I → R does not have the limit L at c if there exists
an > 0 such that the following is true: for every δ > 0 there exists an x ∈ Ṽδ (c) such that
|f (x) − L| > .
Problem 2.) Prove the following limits do not exist
a.)
lim
x→0
1
1
x
Solution: (Let L ∈ R, we will show L is not a limit. Since it is arbitrary we will
conclude there is no limit.) Let L ∈ R be fixed and choose = L2 . Given any δ > 0,
1
1
pick x = min{ 2δ , 2L
}. Both 2δ and 2L
are positive so it must be that x >. Also,
δ
1
0 < |x| ≤ 2 < δ so x ∈ Ṽδ (0). On the other hand, x ≤ 2L
so 2L ≤ x1 and using the
triangle inequality we see
|f (x) − L| = |
1
1
L
− L| ≥ || | − L| ≥ 2L − L = L > = x
x
2
b.)
lim f (x)
x→0
where f (x) = 1 if x is rational and f (x) = 0 if x is irrational.
Solution First we show 0 is not a limit. Let = 12 . Given any δ > 0 let x ∈ Ṽδ (0) be
such that x is rational. We know such an x exists by the density of rational numbers.
For this x, |f (x) − 0| = 1 > 12 = .
Now we show L 6= 0 is not a limit. Let = |L|
2 . Given any δ > 0 let x ∈ Ṽδ (0) be
such that x is irrational. We know such an x exists by the density of irrational
numbers and for this x,
|f (x) − L| = |0 − L| >
|L|
= .
2
c.)
lim f (x)
x→2
where f (x) = (x − 2)2 if x ≤ 2 and f (x) = (x + 2) if x > 2.
Solution First we show 0 is not a limit. Let = 1. Given any δ > 0 let x ∈ Ṽδ (2) be
such that x > 2. Then
|f (x) − 0| = |x + 2| > 4 > Next we show L 6= 0 is not a limit. Let = |L|
2 . Given any δ > 0 let x ∈ Ṽδ (2) be
p
such that 2 − L/2 < x < 2. Then, |(x − 2)2 < L/2 and
|f (x) − L| = |(x − 2)2 − L| ≥ ||(x − 2)2 | − L| >
L
=
2
Note: All of these problems can be solved (maybe easier) by consider sequences for which
the image of the sequences have different limits.
Problem 3.)
Definition 0.2. We say f : I → R is bounded in a neighborhood of c ∈ I if there exists a
δ > 0 and a M > 0 such that if |x − c| < δ then |f (x)| < M .
a.) Prove the following statement: If f is bounded in a neighborhood of c then there
exists a sequence xn → c and a number L ∈ R such that
lim f (xn ) = L
n→∞
2
(Hint: use the Bolzano-Weierstrass theorem).
Solution: Define the sequence xn = c − n1 and assume f is bounded in the δ
neighborhood of c. This sequence converges to c so there exists an Nδ > such that if
n > Nδ then |xn − c| < δ. Consider the sequence yn = f (xn ). Since f is bounded in
the δ neighborhood of c, it must be that |yn | = |f (xn )| < M for any n > Nδ . On the
other hand, {yn }n≤Nδ is a finite set so it has a maximum, say M1 . Therefore the
sequence yn is bounded by max{M1 , M } and the Bolzano-Weierstrass theorem yields
the existence of a subsequence yni and a limit L ∈ R such that yni → L. This is
equivalent to
lim f (xni ) = L
n→∞
In addition, xni is a subsequence of xn so it must have the same limit, that is
xni → c. xni is therefore the sequence we are looking for.
b.) Prove the converse of the above statement is false.
Solution: Consider the function defined in problem 2, part b. This function is
bounded everywhere by 1. The sequence xn = n1 is a rational sequence so f (xn ) = 0
and
lim f (xn ) = 0
n→∞
We proved in problem 2 that there is no limit as x → 0 for this function.
Problem 4.)Let f and g be such that
lim f (x) = L
x→c
lim g(x) = M
x→L
Prove
lim g(f (x)) = M
x→c
Solution: Let > 0 be given. Since g(x) → M as x → L, there exists an η > 0 such that
if 0 < |x − L| < η then |g(x) − L| < . Since f (x) → M as x → c, there exists a δ > 0 such
that if 0 < |x − c| < δ then |f (x) − L| < η. This is the δ we want. If 0 < |x − c| < δ then
|f (x) − L| < η and |g(f (x)) − M | < .
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