T HE U NIVERSITY OF SYDNEY
SCHOOL OF M ATHEMATICS AND STATISTICS
Problem Sheet for Week 5
MATH1901: Differential Calculus (Advanced)
Semester 1, 2017
Web Page: sydney.edu.au/science/maths/u/UG/JM/MATH1901/
Lecturer: Daniel Daners
Material covered
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Intuitive concept of a limit.
Formal definition of limits in terms of 𝜀 and 𝛿.
The limit laws.
The squeeze law.
Outcomes
After completing this tutorial you should
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understand the intuitive notion of limits;
be able to work with the 𝜀-𝛿 definition of a limit in concrete and theoretical contexts;
be able to prove the limit laws;
calculate complicated limits using limit laws and the squeeze law;
use the squeeze law to prove theoretical results;
Summary of essential material
Definition of a limit. We say that lim 𝑓 (𝑥) = 𝓁 if for all 𝜀 > 0 there is a 𝛿 > 0 such that
𝑥→𝑎
0 < |𝑥 − 𝑎| < 𝛿
⇐⇒
Note that
lim 𝑓 (𝑥) = 𝓁
𝑥→𝑎
|𝑓 (𝑥) − 𝓁| < 𝜀.
lim |𝑓 (𝑥) − 𝓁| = 0.
⇐⇒
𝑥→𝑎
We often use the latter in conjuction with the squeeze law. We can also consider limits from the right: We say
that lim 𝑓 (𝑥) = 𝓁 if for all 𝜀 > 0 there is a 𝛿 > 0 such that
𝑥→𝑎+
0<𝑥−𝑎<𝛿
⇐⇒
|𝑓 (𝑥) − 𝓁| < 𝜀.
Similarly we consider limits from the left: replace 𝑥 → 𝑎+ by 𝑥 → 𝑎− and 0 < 𝑥 − 𝑎 < 𝛿 by 0 < 𝑎 − 𝑥 < 𝛿.
• A limit exists if and only if right and left hand limits exist and are equal.
Note on computing limits. The 𝜀-𝛿 definition of a limit cannot be used to compute a limit! It can be used to
prove some number is the limit. If you do that you must make sure that the argument you provide works for
every choice of 𝜀 > 0! The limit laws and the squeeze law, in conjunction with a number of elementary limits,
are the main tools to compute limits!
Limit Laws. If lim 𝑓 (𝑥) = 𝓁 and lim 𝑔(𝑥) = 𝑚, then
𝑥→𝑎
𝑥→𝑎
(1) lim (𝑘𝑓 (𝑥)) = 𝑘𝓁 for all 𝑘 ∈ ℝ.
(3) lim (𝑓 (𝑥)𝑔(𝑥)) = 𝓁𝑚.
(2) lim (𝑓 (𝑥) + 𝑔(𝑥)) = 𝓁 + 𝑚.
(4) lim
𝑥→𝑎
𝑥→𝑎
𝑓 (𝑥)
𝓁
= provided 𝑚 ≠ 0.
𝑥→𝑎 𝑔(𝑥)
𝑚
𝑥→𝑎
Squeeze Law.
Suppose that
𝑓 (𝑥) ≤ 𝑔(𝑥) ≤ ℎ(𝑥)
for all 𝑥 near 𝑎 (but not necessarily at 𝑥 = 𝑎).
If lim 𝑓 (𝑥) = lim ℎ(𝑥) = 𝓁 then lim 𝑔(𝑥) = 𝓁.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
Copyright © 2017 The University of Sydney
1
Questions to complete during the tutorial
Questions marked by * are harder questions.
1. Calculate the following limits using one or more of the limit laws and squeeze law. It is very tedious to
write this down, please at least explain your group exactly how to apply the laws.
𝑥2 + 3𝑥 + 2
𝑥→3 4𝑥2 − 𝑥 + 1
𝑥2 − 3𝑥 + 2
𝑥→1
𝑥3 − 1
√
√
3 + 2𝑥 − 3
(d) lim
𝑥→0
𝑥
(a) lim
(c) lim
𝑥2 − 1
𝑥→1 𝑥 − 1
(b) lim
1
2. Use the Squeeze Law to calculate the limit lim 𝑥2 sin .
𝑥→0
𝑥
3. Sketch the function with formula
⎧1
⎪
𝑓 (𝑥) = ⎨0
⎪2𝑥 + 1
⎩
if 𝑥 < 0,
if 𝑥 = 0,
if 𝑥 > 0.
Find suitable values of 𝛿 such that whenever 0 < |𝑥| < 𝛿, we have:
(a) |𝑓 (𝑥) − 1| < 0.01
4.
(b) |𝑓 (𝑥) − 1| < 0.001
(c) |𝑓 (𝑥) − 1| < 𝜀.
√
√
(a) Let 𝑎 > 0. Use the squeeze law to show that lim 𝑥 = 𝑎.
𝑥→𝑎
√
√
Hint: Rewrite | 𝑥 − 𝑎| so that |𝑥 − 𝑎| appears in the expression.
√
(b) Use the 𝜀-𝛿 definition of limits to show that lim𝑥→0+ 𝑥 = 0.
5. The function 𝑓 is defined by
{
𝑓 (𝑥) =
𝑥2
0
if 𝑥 is rational,
if 𝑥 is irrational.
Use the squeeze law to show that lim𝑥→0 𝑓 (𝑥) = 0. Also write down an 𝜀-𝛿 proof.
6. Suppose that lim 𝑓 (𝑥) = 𝓁 and lim 𝑔(𝑥) = 𝑚. Use the 𝜀-𝛿 definition of limits to prove the following
𝑥→𝑎
𝑥→𝑎
limit laws:
(a) lim (𝑘𝑓 (𝑥)) = 𝑘𝓁, 𝑘 ∈ ℝ is a constant.
(b) lim (𝑓 (𝑥) + 𝑔(𝑥)) = 𝓁 + 𝑚.
𝑥→𝑎
7.
𝑥→𝑎
(a) Assume that lim 𝑓 (𝑥) = 𝓁. Let 𝑀1 , 𝑀2 such that 𝑀1 < 𝓁 < 𝑀2 . Use the 𝜀-𝛿 definition of a
𝑥→𝑎
limit to show that there exists a 𝛿 > 0 such that 𝑀1 < 𝑓 (𝑥) < 𝑀2 for all 𝑥 ∈ (𝑎 − 𝛿, 𝑎 + 𝛿), 𝑥 ≠ 𝑎.
(
)
*(b) If lim 𝑓 (𝑥) = 𝓁 and lim 𝑔(𝑥) = 𝑚, show that lim 𝑓 (𝑥)𝑔(𝑥) = 𝓁𝑚 (limit law of multiplication).
𝑥→𝑎
𝑥→𝑎
(
)𝑥→𝑎 (
)
Hint: Write 𝑓 (𝑥)𝑔(𝑥) − 𝓁𝑚 = 𝑓 (𝑥) 𝑔(𝑥) − 𝑚 + 𝑚 𝑓 (𝑥) − 𝓁 and use the 𝜀-𝛿 definition of limits.
8. Prove or disprove the following statements:
(a) If lim𝑥→0 𝑓 (𝑥)2 exists then lim𝑥→0 𝑓 (𝑥) exists.
(b) If lim𝑥→0 𝑓 (𝑥2 ) exists then lim𝑥→0 𝑓 (𝑥) exists.
*(c) If lim 𝑔(𝑥) = 𝑚 and lim 𝑓 (𝑦) = 𝓁 then lim 𝑓 (𝑔(𝑥)) = 𝓁. Here 𝑓 and 𝑔 both have domain ℝ.
𝑥→𝑎
𝑦→𝑚
𝑥→𝑎
2
Extra questions for further practice
9. Prove that lim 𝑓 (𝑥) = 𝓁 if and only if lim |𝑓 (𝑥) − 𝓁| = 0.
𝑥→𝑎
10.
𝑥→𝑎
|
|
|
|
(a) For 𝑥, 𝑦 ∈ ℝ we obviously have |𝑥| = |(𝑥 − 𝑦) + 𝑦| and |𝑦| = |(𝑦 − 𝑥) + 𝑥|. Use the triangle
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inequality to show that ||𝑥| − |𝑦|| ≤ |𝑥 − 𝑦|. This is called the reversed triangle inequalty.
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(b) Hence, use the squeeze law to show lim 𝑓 (𝑥) = 𝓁 implies that lim |𝑓 (𝑥)| = |𝓁|.
𝑥→𝑎
𝑥→𝑎
(c) Is the converse of the statement in part (b) true? Give a proof or a counter example.
11. Prove the following results using the 𝜖, 𝛿 definition:
{
5 − 2𝑥 if 𝑥 ≠ 4,
(a) lim𝑥→4 𝑓 (𝑥) = −3, where 𝑓 (𝑥) =
100
if 𝑥 = 4.
{
3𝑥 if 𝑥 is rational,
(b) lim𝑥→0 𝑔(𝑥) = 0, where 𝑔(𝑥) =
7𝑥 if 𝑥 is irrational.
12. Sketch the graph of the function:
⎧0
⎪
𝑓 (𝑥) = ⎨1
⎪𝑥 + 2
⎩
𝑥 < 0,
𝑥 = 0,
𝑥 > 0.
Find lim𝑥→0− 𝑓 (𝑥) and lim𝑥→0+ 𝑓 (𝑥) (no need for formal proofs). Does lim𝑥→0 𝑓 (𝑥) exist?
13. Using the limit laws, show that the limit of a function exists as 𝑥 → 𝑎, then the limit is unique. That is,
prove that if lim𝑥→𝑎 𝑓 (𝑥) = 𝓁 and lim𝑥→𝑎 𝑓 (𝑥) = 𝑚, then 𝓁 = 𝑚.
Hint: Write 𝓁 − 𝑚 = (𝑓 (𝑥) − 𝑚) − (𝑓 (𝑥) − 𝓁).
Challenge questions (optional)
{
14. Prove that lim 𝑓 (𝑥) does not exist, where 𝑓 (𝑥) =
𝑥→0
0 if 𝑥 rational,
1 if 𝑥 is irrational.
*15. Prove the quotient limit law: If lim 𝑔(𝑥) = 𝑚 and 𝑚 ≠ 0, then lim
𝑥→𝑎
𝑥→𝑎
3
1
1
= .
𝑔(𝑥) 𝑚