December 9, 2004
Lecturer: Dr Martin Kurth
Michaelmas Term 2004
Problem Sheet 3 – Solutions
1. The function f approaches the limit
lim f (x) = L
x→x0
if for every real number ² > 0 there is a real number δ > 0 such that for
all x in the domain of f
0 < |x − x0 | < δ ⇒ |f (x) − L| < ².
2. For x 6= 3, the first function can be rewritten as
f (x) =
(x + 3)(x − 3)
= x + 3,
x−3
and thus
lim f (x) = lim (x + 3) = 3 + 3 = 6.
x→3
x→3
The second and third functions haven’t got a limit as they “grow too
large” as x approaches 3. The fourth function “jumps” at x = 3, and thus
has no limit.
3. This is more complicated. The answer is that the function f does not
have a limit at x=0. To prove this, we first prove some lemmas (auxiliary
theorems).
• Lemma 1: If an integer number m is even, m2 is even.
Proof: As m is even, there exists an integer number k with
m = 2k,
and thus
m2 = 4k 2 ,
and as k 2 is an integer number, m2 is even.
• Lemma 2: If an integer number n is odd, n2 is odd.
Proof: As n is odd, n + 1 is even, so we know from Lemma 1 that
(n + 1)2 is even. Now we have
(n + 1)2 = n2 + 2n + 1,
and thus
n2 = (n + 1)2 − 2n + 1.
As we know that (n + 1)2 is even, and 2n is even by definition,
(n + 1)2 − 2n is even, so n2 must be odd.
1
√
• Lemma 3: 2 is an irrational number (ie not a√rational number).
Proof (by contradiction): Let us assume that 2 is rational. Then
√
2 can be written as
√
m
2=
n
with m and n integer numbers that haven’t got a common divisor (if
they have, just cancel the common divisor, so we end up with m and
n that haven’t got one). Now we write
√
m2
( 2)2 = 2 = 2 ,
n
ie
m2 = 2n2 .
This means that m2 is even, and thus m must be even (if it was
odd, then m2 would be odd by Lemma 2), so there exists an integer
number k, such that
m = 2k,
which means
m2 = 4k 2 .
This gives us
4k 2 = 2n2 ,
which means
n2 = 2k 2 ,
ie n2 is even, and with the same argument as for m we find out that
n is even. We have now shown that m and n are even, ie they have 2
as a common divisor, contrary to the assumption that they have
√ no
2 is
common divisor at all. This
means
that
the
assumption
that
√
rational is wrong, hence 2 must be irrational.
• Lemma 4: For every real number δ > 0 there exists a rational number
x with
|x| < δ.
Proof: Choose an integer number q with
q>
1
.
δ
x=
1
q
Then
is rational and satisfies
|x| =
as q > 1/δ.
2
1
<δ
q
• Lemma 5: For every real number δ > 0 there exists an irrational
number z with
|z| < δ.
Proof: From Lemma 4 we know that there is a rational number x of
the form x = 1/q with integer q and
|x| < δ.
Now consider
1
z= √ .
q 2
First we have to show that z is irrational. We do this by contradiction, so let us assume z is rational, ie there are integer numbers m
and n with
m
z= .
n
This immediately gives us
√
2=
n
,
mq
√
√
ie 2 is rational. As we know from Lemma 3 that 2 is not rational,
our √
assumption must be wrong, and z must be irrational.
As 2 > 1 we also know that
1
1
|z| = √ < = |x| < δ,
q
q 2
which proves Lemma 5.
Now we can proceed to prove our theorem that f has no limit at x = 0.
Again, let us assume the opposite, ie that such a limit exists.
• Assumption 1:
lim = 1.
x→0
Let us choose ² = 1/2. Now we must find a δ > 0, such that for all x
0 < |x| < δ ⇒ |f (x) − 1| < ².
But from Lemma 5 we know that for every δ we choose, there will
be a number z with |z| < δ that is irrational, ie f (z) = 0. In other
words:
|f (z) − 1| = |0 − 1| = 1 > ².
This means that 1 can’t be the limit of f (x) at x = 0.
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• Assumption 2:
lim = L 6= 1.
x→0
Now let us choose
1
|1 − L|.
2
Then, for every δ we choose, we know from Lemma 4 that there is a
number x with |x| < δ which is rational, ie
²=
f (x) = 1.
We thus have
|f (x) − L| = |1 − L| > ²,
which means that L 6= 1 can’t be the limit of f (x) at x = 0 either.
This means that f (x) has no limit as x approaches 0.
Note: Don’t believe that mathematicians think in this structured way, ie
they sit down, and think “I’ll start with 5 lemmas, and Lemma 1 will be
. . . ”. They will normally start to scribble down the proof of the “big”
theorem, and then notice what lemmas they need as they go along. When
they write their proof down properly, it looks as well planned as this page,
but a mathematician’s scratch paper doesn’t look anything like this!
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