MA2002: Tutorial
He Taoran
1
Tutorial 1 (Week 3)
Review
1. Composite functions: f ◦ g(x) := f (g(x)), which is well-defined if and
only if x ∈ Dom(g) and g(x) ∈ Dom(f ). Here Dom(f ) denotes the domain of
a function f . Note that in general f ◦ g ̸= g ◦ f . Inductively, one can define the
composition of n functions f1 , f2 , · · · , fn :
fn ◦ fn−1 ◦ · · · ◦ f1 (x) := fn (fn−1 ◦ · · · ◦ f1 (x))
with n ≥ 3.
2. Odd and even functions: Given a function f whose domain Dom(f ) is
symmetric about 0, we say
f
f
is even
⇐⇒
f (x) = f (−x)
is odd
⇐⇒
f (x) = −f (−x)
for any x ∈ Dom(f ),
for any x ∈ Dom(f ).
If f is odd and 0 ∈ Dom(f ), we immediately have f (0) = 0. However, lim f (x)
x→0
may not exist. If it exists, one can prove that lim f (x) must be zero (why ?).
x→0
Note that in order to show that a function f is neither even nor odd, one needs
to find x1 , x2 ∈ Dom(f ) such that f (x1 ) ̸= f (−x1 ) and f (x2 ) ̸= −f (−x2 ).
3. Limit laws: Suppose lim f (x) = L and lim g(x) = M , then there hold
x→a
lim (f (x) + g(x)) = L + M,
x→a
lim (f (x)g(x)) = LM,
x→a
x→a
lim (cf (x)) = cL with c being a constant,
f (x)
L
lim
=
provided that M ̸= 0.
x→a
g(x)
M
x→a
To evaluate lim f (x)/g(x) when M = lim g(x) = 0, we consider two cases as
x→a
x→a
below:
• Case 1: If L ̸= 0, then lim f (x)/g(x) does not exist.
x→a
• Case 2: If L = 0, we try to find the common factor of f and g in the form
of (x − a)k for some k ∈ Z≥1 . Suppose that we have the factorization
f (x) = (x − a)k f1 (x)
and g(x) = (x − a)k g1 (x),
1
where lim f1 (x) exists (which is not necessarily non-zero) and lim g1 (x) ̸=
x→a
x→a
0, then
lim f1 (x)
f (x)
f1 (x)
lim
= lim
= x→a
.
x→a g(x)
x→a g1 (x)
lim g1 (x)
x→a
As a corollary of the discussion above, we also obtain that if lim f (x)/g(x)
x→a
exists and lim g(x) = 0, then it must hold that lim f (x) = 0.
x→a
x→a
4. Squeeze theorem: Let f, g, h be functions satisfying g(x) ≤ f (x) ≤ h(x)
for all x near a (except at a). If lim g(x) = lim h(x) = L, then lim f (x) exists
x→a
x→a
x→a
and equals to L. Note that the squeeze theorem also holds true for one-sided
limit.
Further exercise
Question 1. For f (x) =
of f ◦ g ◦ h.
2
x+1 , g(x)
= cos x and h(x) =
√
x + 3, find the domain
Answer. Dom(f ◦ g ◦ h) = [−3, ∞)\{(2k + 1)2 π 2 − 3 : k ∈ Z}.
√
√
√ to express
Question 2. Let a, b > 0. Mimic the formula a − b = √a−b
a+ b
√
√
3
a−b
3
a − b in the form of □ .
hint. Use the identity x3 − y 3 = (x − y)(x2 + xy + y 2 ).
Answer.
√
3
a−
√
3
b=
a−b
2
3
1
1
2
a + a3 b3 + b3
.
1
: k ∈ Z≥1 }. Determine the infinite limit
Question 3. Let D = R+ \{ kπ
lim
x→0+ , x∈D
1
.
x sin(1/x)
Answer. Neither ∞ nor −∞. Picking xn =
n ≥ 1, then
1
(2n+ 12 )π
and yn =
1
1
= lim (2n + )π = ∞,
xn sin(1/xn ) n→∞
2
1
1
lim
= lim −(2n − )π = −∞.
n→∞ yn sin(1/yn )
n→∞
2
lim
n→∞
2
1
(2n− 21 )π
for
2
Tutorial 2 (Week 4)
Review
1. Precise definition of limits:
• lim f (x) = L if and only if for any ϵ > 0, there exists δ > 0 so that for all
x→a
0 < |x − a| < δ, it holds |f (x) − L| < ϵ.
• lim+ f (x) = L if and only if for any ϵ > 0, there exists δ > 0 so that for
x→a
all a < x < a + δ, it holds |f (x) − L| < ϵ.
• lim f (x) = ∞ if and only if for any M > 0, there exists δ > 0 so that for
x→a
all 0 < |x − a| < δ, it holds f (x) > M .
• lim f (x) = ∞ if and only if for any M > 0, there exists N > 0 so that
x→∞
for all x > N , it holds f (x) > M .
Remark. Here ϵ > 0 can be chosen arbitrarily. However, once a specific ϵ > 0
is given, we must treat ϵ as a fixed number and seek for a suitable δ > 0. In
general, the choice of δ will depend on ϵ, a and also the function f (x). Note that
it is acceptable that we only obtain |f (x) − L| ≤ Cϵ, as we can set the “new”
ϵ′ = (C + 1)ϵ.
When we want to prove lim f (x) = L utilizing the definition of limit, we have
x→a
the freedom to choose ϵ, δ (or N, M ) at the beginning. For example, we can assume that ϵ, δ ∈ (0, 1) (or N, M > 1). This will give some additional constraints
for x ∈ (a − δ, a + δ)\{a} (or x > N ), which may be helpful in finding δ > 0.
2. Strategy of proving lim f (x) = L by ϵ, δ-definition of limit: The main
x→a
task is to find a suitable δ > 0 such that
0 < |x − a| < δ
=⇒
|f (x) − L| < ϵ.
• Step 1: Estimate |f (x) − L| ≤ g(x, a)|x − a| for certain non-negative
function g(x, a).
• Step 2: Find an upper bound for g(x, a). You may rewrite g(x, a) in terms
of x − a, and prove that g(x, a) ≤ M (δ) for any 0 < |x − a| < δ.
• Step 3: If M (δ) ≤ M ′ for all 0 < δ ≤ δ0 with M ′ > 0, δ0 > 0, then pick
δ = min{δ0 , ϵ/M ′ }. Thus
0 < |x − a| < δ
=⇒
|f (x) − L| ≤ g(x, a)|x − a| < M ′ δ ≤ ϵ.
Remark. Generally, it is hard to determine the set A = {x : |f (x) − L| < ϵ}
explicitly. The aim of this strategy is to find a simpler upper bound of |f (x)−L|,
i.e. M ′ |x − a|. Denoting B = {0 < |x − a| < δ ≤ δ0 : M ′ |x − a| < ϵ}, then B is
a subset of A (why?). Thus, it suffices to find δ > 0 such that
0 < |x − a| < δ
=⇒
3
x ∈ B ⊂ A.
If M (δ) is of size δ −α with α ∈ (0, 1), we may fail to find a favorable M ′ in Step
3. Fortunately, we can still bound
|f (x) − L| ≤ Cδ −α |x − a| < Cδ 1−α
0 < |x − a| < δ.
for all
1
Here C > 0 is a constant. Hence we √
can set δ = (ϵ/C) 1−α > 0. You can apply
this modified approach to prove lim x = 0.
x→0
3. Triangle inequality: For any a, b ∈ R, it holds |a| − |b| ≤ |a + b| ≤ |a| + |b|.
Interchanging a and b, we also obtain |b|−|a| ≤ |a+b|. Merging these inequalities
together, we have
|a| − |b| ≤ |a + b| ≤ |a| + |b|.
Further exercise
Question 1. Prove that |a + b + c| ≤ |a| + |b| + |c| for any a, b, c ∈ R.
Answer. |a + b + c| = |(a + b) + c| ≤ |a + b| + |c| ≤ |a| + |b| + |c|.
Question 2. Find lim+ ⌊x + 1⌋. Use this result to identify whether lim ⌊x + 1⌋
x→9
x→9
exists or not.
Answer. For any 9 < x < 10, 10 < x + 1 < 11, thus ⌊x + 1⌋ = 10, which
implies lim ⌊x + 1⌋ = 10. Since lim ⌊x + 1⌋ = 10 ̸= 9 = lim ⌊x + 1⌋, the limit
x→9−
x→9+
x→9+
lim ⌊x + 1⌋ does not exist.
x→9
Question 3. Let fn (x) = nx with n ∈ Z≥1 . It is easy to see that lim fn (x) = 0
x→0
for all n ≥ 1. Pick ϵ > 0.
(a) Fix n ≥ 1. Determine all δn > 0 such that
0 < |x| < δn
=⇒
|fi (x)| < ϵ for any 1 ≤ i ≤ n.
(b) Can you find a unified constant δ > 0 such that
0 < |x| < δ
Answer. (a) Take δn =
ϵ
n.
=⇒
|fn (x)| < ϵ for all n ≥ 1?
Then for any 0 < |x| < δn and 1 ≤ i ≤ n, we have
|fi (x)| = i|x| ≤ n|x| < nδn = ϵ.
(b) No. Suppose such δ > 0 exists. Choosing x =
(why?), we obtain
nδ
ϵ > |fn (x)| =
> ϵ,
2
a contradiction!
4
1
2δ
and n = ⌊ 2ϵ
δ ⌋+1 >
2ϵ
δ
3
Tutorial 3 (Week 5)
Review
1. Continuity: f is continuous at x = a if and only if lim f (x) = f (a).
x→a
• Require that a ∈ Dom(f ), lim f (x) exists and equals to f (a).
x→a
• (By Tut 2 Q4)
f is continuous at x = a ⇐⇒ lim+ f (x) = lim− f (x) = f (a).
x→a
x→a
2. Discontinuity: f is not continuous at x = a if and only if one of following
scenarios occurs:
(a) a ∈
/ Dom(f );
(b) lim f (x) does not exist (jump/infinite/oscillatory discontinuity);
x→a
(c) lim f (x) ̸= f (a) (removable discontinuity).
x→a
3. Intermediate value theorem:
If f is continuous in [a, b] and f (a) ̸= f (b). Then for any N between f (a) and
f (b), there exists c ∈ (a, b) such that f (c) = N .
Corollary (Locations of roots theorem):
If f is continuous in [a, b] and f (a)f (b) < 0. Then we can find c ∈ (a, b) such
that f (c) = 0.
4. Derivative:
f ′ (a) := lim
x→a
f (x) − f (a)
.
x−a
f is differentiable at x = a if and only if f ′ (a) exists.
• Differentiability implies continuity, i.e., f ′ (a) exists =⇒ lim f (x) = f (a).
x→a
However, f ′ (x) may not be continuous.
• Geometric meaning: slope of the tangent line for y = f (x) at (a, f (a)).
Equation of the tangent line: y = f ′ (a)(x − a) + f (a).
5. Differentiation Formulas:
(f + g)′ = f ′ + g ′ ,
(cf )′ = cf ′ (c is a constant),
f
f ′ g − f g′
( )′ =
,
g
g2
Chain rule: (f ◦ g)′ (a) = f ′ (g(a))g ′ (a),
(f g)′ = f ′ g + f g ′ ,
(xα )′ = αxα−1 ,
(sin x)′ = cos x,
5
(cos x)′ = − sin x.
Further exercise
Question 1. Is f (x) = |x + 2| differentiable at x ̸= −2? Find f ′ (a) for a ̸= −2.
Question 2. Let f (x), g(x) be two differentiable functions. Define
(
f (x)
if x ≤ a,
h(x) =
g(x)
if x > a.
Determine the conditions that make f differentiable.
6