Complex Functions : Limits and continuity Ananda Dasgupta MA211, Lecture 6
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Complex Functions :Limits and continuity
Ananda Dasgupta
MA211, Lecture 6
Continuous functions on RThe intuitive notion :
A function whose graph can be drawn withoutlifting pen from paper.
I Easy to understand.
I Hard to use in rigorous discussions.
x
f (x)
Continuous functions on RThe intuitive notion :
A function whose graph can be drawn withoutlifting pen from paper.
I Easy to understand.
I Hard to use in rigorous discussions.
x
f (x)
Continuous functions on RThe intuitive notion :
A function whose graph can be drawn withoutlifting pen from paper.
I Easy to understand.
I Hard to use in rigorous discussions.
x
f (x)
Continuous functions on RPrecise definition :
A function f : S ⊂ R → R is continuous at x0 ∈ Sif ∀ε > 0, ∃δ > 0 :
|x − x0| < δ =⇒ |f (x)− f (x0)| < ε
x
f (x)
x0
f (x0)
Continuous functions on RPrecise definition :
A function f : S ⊂ R → R is continuous at x0 ∈ Sif ∀ε > 0, ∃δ > 0 :
|x − x0| < δ =⇒ |f (x)− f (x0)| < ε
We can prove rigorous results with this.
x
f (x)
x0
f (x0)
Continuous functions on RGeometric meaning
No matter how finicky we are, we can confine the valueof f (x) to a narrow enough band centered at f (x0),
by keeping x confined to a sufficiently narrow bandcentered at x0.
x
f (x)
x0
f (x0)2ε
Continuous functions on RGeometric meaning
No matter how finicky we are, we can confine the valueof f (x) to a narrow enough band centered at f (x0),by keeping x confined to a sufficiently narrow bandcentered at x0.
x
f (x)
x0
f (x0)2ε
2δ
Continuous functions on RGeometric meaning
No matter how finicky we are, we can confine the valueof f (x) to a narrow enough band centered at f (x0),
by keeping x confined to a sufficiently narrow bandcentered at x0.
x
f (x)
x0
f (x0)2ε
2δ
Continuous functions on RGeometric meaning - a different look
We can picture map f : x 7→ f (x) as carrying pointsin one copy of the real line to another.
Continuous functions on RGeometric meaning - a different look
x0 f (x0)
It carries the point at x0 to the point at f (x0).
Continuous functions on RGeometric meaning - a different look
x0 f (x0)
2ε
It carries the point at x0 to the point at f (x0).Consider the interval (f (x0)− ε, f (x0) + ε) of width2ε centered around f (x0).
Continuous functions on RGeometric meaning - a different look
x0 f (x0)
2ε2δ
It carries the point at x0 to the point at f (x0).Consider the interval (f (x0)− ε, f (x0) + ε) of width2ε centered around f (x0).The definition of continuity means that we can alwaysfind a sufficiently small open interval centered at x0
Continuous functions on RGeometric meaning - a different look
x0 f (x0)
2ε2δ
It carries the point at x0 to the point at f (x0).Consider the interval (f (x0)− ε, f (x0) + ε) of width2ε centered around f (x0).The definition of continuity means that we can al-ways find a sufficiently small open interval cen-tered at x0 so that f carries it inside the interval(f (x0)− ε, f (x0) + ε).
Limits on the real line
If f : S ⊂ R→ R is defined in a neighbourhood ofx0, except possibly at x0, then it has a limit a atx0 ∈ S if ∀ε > 0, ∃δ > 0 :
0 < |x − x0| < δ =⇒ |f (x)− a| < ε
We writelimx→x0
f (x) = a
A function f is continuous at x0 iff :
I f (x0) exists.I limx→x0
f (x) exists.I limx→x0
f (x) = f (x0).
Limits on the real line
If f : S ⊂ R→ R is defined in a neighbourhood ofx0, except possibly at x0, then it has a limit a atx0 ∈ S if ∀ε > 0, ∃δ > 0 :
0 < |x − x0| < δ =⇒ |f (x)− a| < ε
We writelimx→x0
f (x) = a
A function f is continuous at x0 iff :
I f (x0) exists.I limx→x0
f (x) exists.I limx→x0
f (x) = f (x0).
Distance on the real line
The geometric viewpoint stresses the importance ofopen intervals, and hence, distance in the discussionof limits and continuity.
The distance between two points x , y ∈ R is
|x − y | =
{x − y for x ≥ y
y − x for x < y
It has the following properties
I |x − y | ≥ 0 with equality holding iff x = y .
I |x − y | = |y − x |.I |x − z | ≤ |x − y |+ |y − z |.
Distance on the real line
The geometric viewpoint stresses the importance ofopen intervals, and hence, distance in the discussionof limits and continuity.The distance between two points x , y ∈ R is
|x − y | =
{x − y for x ≥ y
y − x for x < y
It has the following properties
I |x − y | ≥ 0 with equality holding iff x = y .
I |x − y | = |y − x |.I |x − z | ≤ |x − y |+ |y − z |.
Distance on the real line
The geometric viewpoint stresses the importance ofopen intervals, and hence, distance in the discussionof limits and continuity.The distance between two points x , y ∈ R is
|x − y | =
{x − y for x ≥ y
y − x for x < y
It has the following properties
I |x − y | ≥ 0 with equality holding iff x = y .
I |x − y | = |y − x |.I |x − z | ≤ |x − y |+ |y − z |.
Distance on the real line
The geometric viewpoint stresses the importance ofopen intervals, and hence, distance in the discussionof limits and continuity.The distance between two points x , y ∈ R is
|x − y | =
{x − y for x ≥ y
y − x for x < y
It has the following properties
I |x − y | ≥ 0 with equality holding iff x = y .
I |x − y | = |y − x |.I |x − z | ≤ |x − y |+ |y − z |.
Distance on the real line
The geometric viewpoint stresses the importance ofopen intervals, and hence, distance in the discussionof limits and continuity.The distance between two points x , y ∈ R is
|x − y | =
{x − y for x ≥ y
y − x for x < y
It has the following properties
I |x − y | ≥ 0 with equality holding iff x = y .
I |x − y | = |y − x |.
I |x − z | ≤ |x − y |+ |y − z |.
Distance on the real line
The geometric viewpoint stresses the importance ofopen intervals, and hence, distance in the discussionof limits and continuity.The distance between two points x , y ∈ R is
|x − y | =
{x − y for x ≥ y
y − x for x < y
It has the following properties
I |x − y | ≥ 0 with equality holding iff x = y .
I |x − y | = |y − x |.I |x − z | ≤ |x − y |+ |y − z |.
Using the precise definition - an example
Theorem
If the functions f : S ⊂ R→ R and g : S → Rare both continuous at x = x0 ∈ S, then so is thefunction f + g : S → R.
Proof.
• For a given ε, we can find δ1, δ2 > 0 such that
|x − x0| < δ1 =⇒ |f (x)− f (x0)| < ε
2
and |x − x0| < δ2 =⇒ |g(x)− g (x0)| < ε
2
• Choose δ = min {δ1, δ2}.
Using the precise definition - an example
Theorem
If the functions f : S ⊂ R→ R and g : S → Rare both continuous at x = x0 ∈ S, then so is thefunction f + g : S → R.
Proof.
• For a given ε, we can find δ1, δ2 > 0 such that
|x − x0| < δ1 =⇒ |f (x)− f (x0)| < ε
2
and |x − x0| < δ2 =⇒ |g(x)− g (x0)| < ε
2
• Choose δ = min {δ1, δ2}.
Using the precise definition - an example
Theorem
If the functions f : S ⊂ R→ R and g : S → Rare both continuous at x = x0 ∈ S, then so is thefunction f + g : S → R.
Proof.
• For a given ε, we can find δ1, δ2 > 0 such that
|x − x0| < δ1 =⇒ |f (x)− f (x0)| < ε
2
and |x − x0| < δ2 =⇒ |g(x)− g (x0)| < ε
2
• Choose δ = min {δ1, δ2}.
Using the precise definition - an example
Theorem
If the functions f : S ⊂ R→ R and g : S → Rare both continuous at x = x0 ∈ S, then so is thefunction f + g : S → R.
Proof.
• For a given ε, we can find δ1, δ2 > 0 such that
|x − x0| < δ1 =⇒ |f (x)− f (x0)| < ε
2
and |x − x0| < δ2 =⇒ |g(x)− g (x0)| < ε
2
• Choose δ = min {δ1, δ2}.
Using the precise definition - an example
Theorem
If the functions f : S ⊂ R→ R and g : S → Rare both continuous at x = x0 ∈ S, then so is thefunction f + g : S → R.
Proof.
• For a given ε, we can find δ1, δ2 > 0 such that
|x − x0| < δ1 =⇒ |f (x)− f (x0)| < ε
2
and |x − x0| < δ2 =⇒ |g(x)− g (x0)| < ε
2
• Choose δ = min {δ1, δ2}.
Using the precise definition - an example
Theorem
If the functions f : S ⊂ R→ R and g : S → Rare both continuous at x = x0 ∈ S, then so is thefunction f + g : S → R.
Proof.
• For a given ε, we can find δ1, δ2 > 0 such that
|x − x0| < δ1 =⇒ |f (x)− f (x0)| < ε
2
and |x − x0| < δ2 =⇒ |g(x)− g (x0)| < ε
2
• Choose δ = min {δ1, δ2}.
Using the precise definition - the example continued
• Then |x − x0| < δ =⇒
|(f (x) + g(x))− (f (x0) + g (x0))|= |(f (x)− f (x0)) + (g(x)− g (x0))|≤ |f (x)− f (x0)|+ |g(x)− g (x0)|<ε
2+ε
2= ε.
�• What did we use in this proof?• The inequality :
|x − z | ≤ |x − y |+ |y − z |.
Using the precise definition - the example continued
• Then |x − x0| < δ =⇒
|(f (x) + g(x))− (f (x0) + g (x0))|
= |(f (x)− f (x0)) + (g(x)− g (x0))|≤ |f (x)− f (x0)|+ |g(x)− g (x0)|<ε
2+ε
2= ε.
�• What did we use in this proof?• The inequality :
|x − z | ≤ |x − y |+ |y − z |.
Using the precise definition - the example continued
• Then |x − x0| < δ =⇒
|(f (x) + g(x))− (f (x0) + g (x0))|= |(f (x)− f (x0)) + (g(x)− g (x0))|
≤ |f (x)− f (x0)|+ |g(x)− g (x0)|<ε
2+ε
2= ε.
�• What did we use in this proof?• The inequality :
|x − z | ≤ |x − y |+ |y − z |.
Using the precise definition - the example continued
• Then |x − x0| < δ =⇒
|(f (x) + g(x))− (f (x0) + g (x0))|= |(f (x)− f (x0)) + (g(x)− g (x0))|≤ |f (x)− f (x0)|+ |g(x)− g (x0)|
<ε
2+ε
2= ε.
�• What did we use in this proof?• The inequality :
|x − z | ≤ |x − y |+ |y − z |.
Using the precise definition - the example continued
• Then |x − x0| < δ =⇒
|(f (x) + g(x))− (f (x0) + g (x0))|= |(f (x)− f (x0)) + (g(x)− g (x0))|≤ |f (x)− f (x0)|+ |g(x)− g (x0)|<ε
2+ε
2
= ε.
�• What did we use in this proof?• The inequality :
|x − z | ≤ |x − y |+ |y − z |.
Using the precise definition - the example continued
• Then |x − x0| < δ =⇒
|(f (x) + g(x))− (f (x0) + g (x0))|= |(f (x)− f (x0)) + (g(x)− g (x0))|≤ |f (x)− f (x0)|+ |g(x)− g (x0)|<ε
2+ε
2= ε.
�
• What did we use in this proof?• The inequality :
|x − z | ≤ |x − y |+ |y − z |.
Using the precise definition - the example continued
• Then |x − x0| < δ =⇒
|(f (x) + g(x))− (f (x0) + g (x0))|= |(f (x)− f (x0)) + (g(x)− g (x0))|≤ |f (x)− f (x0)|+ |g(x)− g (x0)|<ε
2+ε
2= ε.
�• What did we use in this proof?
• The inequality :
|x − z | ≤ |x − y |+ |y − z |.
Using the precise definition - the example continued
• Then |x − x0| < δ =⇒
|(f (x) + g(x))− (f (x0) + g (x0))|= |(f (x)− f (x0)) + (g(x)− g (x0))|≤ |f (x)− f (x0)|+ |g(x)− g (x0)|<ε
2+ε
2= ε.
�• What did we use in this proof?• The inequality :
|x − z | ≤ |x − y |+ |y − z |.
Continuity beyond the real line
We can introduce the notion of continuity of anfunction between arbitrary sets U and V if there isa notion of distance on the two sets.
The distance function (formally called the metric)on a set U is a map d : U × U → R+ with thefollowing properties :
I d(x , y) ≥ 0 with equality holding iff x = y .
I d(x , y) = d(y , x).
I d(x , z) ≤ d(x , y) + d(y , z).
The ordered pair (U , d) where d is a metricfunction defined on U is called a metric space.
Continuity beyond the real line
We can introduce the notion of continuity of anfunction between arbitrary sets U and V if there isa notion of distance on the two sets.The distance function (formally called the metric)on a set U is a map d : U × U → R+ with thefollowing properties :
I d(x , y) ≥ 0 with equality holding iff x = y .
I d(x , y) = d(y , x).
I d(x , z) ≤ d(x , y) + d(y , z).
The ordered pair (U , d) where d is a metricfunction defined on U is called a metric space.
Continuity beyond the real line
We can introduce the notion of continuity of anfunction between arbitrary sets U and V if there isa notion of distance on the two sets.The distance function (formally called the metric)on a set U is a map d : U × U → R+ with thefollowing properties :
I d(x , y) ≥ 0 with equality holding iff x = y .
I d(x , y) = d(y , x).
I d(x , z) ≤ d(x , y) + d(y , z).
The ordered pair (U , d) where d is a metricfunction defined on U is called a metric space.
Continuity beyond the real line
We can introduce the notion of continuity of anfunction between arbitrary sets U and V if there isa notion of distance on the two sets.The distance function (formally called the metric)on a set U is a map d : U × U → R+ with thefollowing properties :
I d(x , y) ≥ 0 with equality holding iff x = y .
I d(x , y) = d(y , x).
I d(x , z) ≤ d(x , y) + d(y , z).
The ordered pair (U , d) where d is a metricfunction defined on U is called a metric space.
Continuity beyond the real line
We can introduce the notion of continuity of anfunction between arbitrary sets U and V if there isa notion of distance on the two sets.The distance function (formally called the metric)on a set U is a map d : U × U → R+ with thefollowing properties :
I d(x , y) ≥ 0 with equality holding iff x = y .
I d(x , y) = d(y , x).
I d(x , z) ≤ d(x , y) + d(y , z).
The ordered pair (U , d) where d is a metricfunction defined on U is called a metric space.
Continuity beyond the real line
We can introduce the notion of continuity of anfunction between arbitrary sets U and V if there isa notion of distance on the two sets.The distance function (formally called the metric)on a set U is a map d : U × U → R+ with thefollowing properties :
I d(x , y) ≥ 0 with equality holding iff x = y .
I d(x , y) = d(y , x).
I d(x , z) ≤ d(x , y) + d(y , z).
The ordered pair (U , d) where d is a metricfunction defined on U is called a metric space.
Continuity beyond the real line
We can introduce the notion of continuity of anfunction between arbitrary sets U and V if there isa notion of distance on the two sets.The distance function (formally called the metric)on a set U is a map d : U × U → R+ with thefollowing properties :
I d(x , y) ≥ 0 with equality holding iff x = y .
I d(x , y) = d(y , x).
I d(x , z) ≤ d(x , y) + d(y , z).
The ordered pair (U , d) where d is a metricfunction defined on U is called a metric space.
Continuity beyond the real line
We can introduce the notion of continuity of anfunction between arbitrary sets U and V if there isa notion of distance on the two sets.
Given two metric spaces (U , dU) and (V , dV ), afunction f : U → V is defined to be continuous atu0 ∈ U if :∀ε > 0, ∃δ > 0 :
dU (u, u0) < δ =⇒ dV (f (u), f (u0)) < ε.
Continuity beyond the real line
We can introduce the notion of continuity of anfunction between arbitrary sets U and V if there isa notion of distance on the two sets.
Given two metric spaces (U , dU) and (V , dV ), afunction f : U → V is defined to be continuous atu0 ∈ U if :∀ε > 0, ∃δ > 0 :
dU (u, u0) < δ =⇒ dV (f (u), f (u0)) < ε.
R2 as a metric space
I The plane R2 has a natural metric defined on it.
I The Euclidean distance function assigns thepositive real number
d(x , y) =
√(x1 − y1)2 + (x2 − y2)2
to the pair of points x = (x1, x2) andy = (y1, y2).
I It satisfies all the properties of a distancefunction.
R2 as a metric space
I The plane R2 has a natural metric defined on it.
I The Euclidean distance function assigns thepositive real number
d(x , y) =
√(x1 − y1)2 + (x2 − y2)2
to the pair of points x = (x1, x2) andy = (y1, y2).
I It satisfies all the properties of a distancefunction.
R2 as a metric space
I The plane R2 has a natural metric defined on it.
I The Euclidean distance function assigns thepositive real number
d(x , y) =
√(x1 − y1)2 + (x2 − y2)2
to the pair of points x = (x1, x2) andy = (y1, y2).
I It satisfies all the properties of a distancefunction.
Limits and continuity on R2
A function u : S ⊂ R2 → R, (x , y) 7→ u(x , y)defined in some neighbourhood of the point (x0, y0),except possibly at (x0, y0), has the limit u0 at(x0, y0) if ∀ε > 0, ∃δ > 0 :
0 <√
(x−x0)2+(y−y0)
2 < δ =⇒ |u(x , y)− u0| < ε
A function u : S ⊂ R2 → R, (x , y) 7→ u(x , y)defined in some neighbourhood of the point (x0, y0),is continuous at (x0, y0) if ∀ε > 0, ∃δ > 0 :
√(x−x0)
2+(y−y0)2 < δ =⇒ |u(x , y)− u (x0, y0)| < ε
Limits and continuity on R2
A function u : S ⊂ R2 → R, (x , y) 7→ u(x , y)defined in some neighbourhood of the point (x0, y0),except possibly at (x0, y0), has the limit u0 at(x0, y0) if ∀ε > 0, ∃δ > 0 :
0 <√
(x−x0)2+(y−y0)
2 < δ =⇒ |u(x , y)− u0| < ε
A function u : S ⊂ R2 → R, (x , y) 7→ u(x , y)defined in some neighbourhood of the point (x0, y0),is continuous at (x0, y0) if ∀ε > 0, ∃δ > 0 :
√(x−x0)
2+(y−y0)2 < δ =⇒ |u(x , y)− u (x0, y0)| < ε
Examples
lim(x ,y)→(0,0)
u(x , y) = 0, where u(x , y) = x3
x2+y2
Proof.
If x = r cos θ, y = r sin θ, we have
u(x , y) =r 3 cos3 θ
r 2 sin2 θ + r 2 cos2 θ= r cos3 θ
Since√
(x − 0)2 + (y − 0)2 = r , we have
|u(x , y)− 0| = r∣∣cos3 θ
∣∣ < ε
whenever 0 <√
x2 + y 2 = r < ε. Hence, takingδ = ε allows the inequality in the definition of thelimit to be satisfied.
Examples
lim(x ,y)→(0,0)
u(x , y) = 0, where u(x , y) = x3
x2+y2
Proof.
If x = r cos θ, y = r sin θ, we have
u(x , y) =r 3 cos3 θ
r 2 sin2 θ + r 2 cos2 θ= r cos3 θ
Since√
(x − 0)2 + (y − 0)2 = r , we have
|u(x , y)− 0| = r∣∣cos3 θ
∣∣ < ε
whenever 0 <√
x2 + y 2 = r < ε. Hence, takingδ = ε allows the inequality in the definition of thelimit to be satisfied.
Examples
lim(x ,y)→(0,0)
u(x , y) = 0, where u(x , y) = x3
x2+y2
Proof.
If x = r cos θ, y = r sin θ, we have
u(x , y) =r 3 cos3 θ
r 2 sin2 θ + r 2 cos2 θ
= r cos3 θ
Since√
(x − 0)2 + (y − 0)2 = r , we have
|u(x , y)− 0| = r∣∣cos3 θ
∣∣ < ε
whenever 0 <√
x2 + y 2 = r < ε. Hence, takingδ = ε allows the inequality in the definition of thelimit to be satisfied.
Examples
lim(x ,y)→(0,0)
u(x , y) = 0, where u(x , y) = x3
x2+y2
Proof.
If x = r cos θ, y = r sin θ, we have
u(x , y) =r 3 cos3 θ
r 2 sin2 θ + r 2 cos2 θ= r cos3 θ
Since√
(x − 0)2 + (y − 0)2 = r , we have
|u(x , y)− 0| = r∣∣cos3 θ
∣∣ < ε
whenever 0 <√
x2 + y 2 = r < ε. Hence, takingδ = ε allows the inequality in the definition of thelimit to be satisfied.
Examples
lim(x ,y)→(0,0)
u(x , y) = 0, where u(x , y) = x3
x2+y2
Proof.
If x = r cos θ, y = r sin θ, we have
u(x , y) =r 3 cos3 θ
r 2 sin2 θ + r 2 cos2 θ= r cos3 θ
Since√
(x − 0)2 + (y − 0)2 = r , we have
|u(x , y)− 0| = r∣∣cos3 θ
∣∣ < ε
whenever 0 <√
x2 + y 2 = r < ε. Hence, takingδ = ε allows the inequality in the definition of thelimit to be satisfied.
Examples
lim(x ,y)→(0,0)
u(x , y) = 0, where u(x , y) = x3
x2+y2
Proof.
If x = r cos θ, y = r sin θ, we have
u(x , y) =r 3 cos3 θ
r 2 sin2 θ + r 2 cos2 θ= r cos3 θ
Since√
(x − 0)2 + (y − 0)2 = r , we have
|u(x , y)− 0| = r∣∣cos3 θ
∣∣
< ε
whenever 0 <√
x2 + y 2 = r < ε. Hence, takingδ = ε allows the inequality in the definition of thelimit to be satisfied.
Examples
lim(x ,y)→(0,0)
u(x , y) = 0, where u(x , y) = x3
x2+y2
Proof.
If x = r cos θ, y = r sin θ, we have
u(x , y) =r 3 cos3 θ
r 2 sin2 θ + r 2 cos2 θ= r cos3 θ
Since√
(x − 0)2 + (y − 0)2 = r , we have
|u(x , y)− 0| = r∣∣cos3 θ
∣∣ < ε
whenever 0 <√
x2 + y 2 = r < ε.
Hence, takingδ = ε allows the inequality in the definition of thelimit to be satisfied.
Examples
lim(x ,y)→(0,0)
u(x , y) = 0, where u(x , y) = x3
x2+y2
Proof.
If x = r cos θ, y = r sin θ, we have
u(x , y) =r 3 cos3 θ
r 2 sin2 θ + r 2 cos2 θ= r cos3 θ
Since√
(x − 0)2 + (y − 0)2 = r , we have
|u(x , y)− 0| = r∣∣cos3 θ
∣∣ < ε
whenever 0 <√
x2 + y 2 = r < ε. Hence, takingδ = ε allows the inequality in the definition of thelimit to be satisfied.
An important property
If the limit lim(x ,y)→(x0,y0)
u(x , y) = u0 exists, then
u(x , y) must approach u0 as the point (x , y)approaches the point (x0, y0) along any curve.
If we can find two curves C1, C2 going through(x0, y0) along which u(x , y) approaches twodifferent values u1 and u2 as (x , y) approaches(x0, y0), then lim
(x ,y)→(x0,y0)u(x , y) does not exist.
An important property
If the limit lim(x ,y)→(x0,y0)
u(x , y) = u0 exists, then
u(x , y) must approach u0 as the point (x , y)approaches the point (x0, y0) along any curve.
If we can find two curves C1, C2 going through(x0, y0) along which u(x , y) approaches twodifferent values u1 and u2 as (x , y) approaches(x0, y0), then lim
(x ,y)→(x0,y0)u(x , y) does not exist.
Examples
lim(x ,y)→(0,0)
xy
x2 + y 2does not exist
Proof.
If (x , y) approaches (0, 0) along the x-axis, then
lim(x ,0)→(0,0)
u(x , 0) = limx→0
(x)(0)
x2 + 02= 0
However, if (x , y) approaches (0, 0) along the liney = x , then
lim(x ,x)→(0,0)
u(x , x) = limx→0
(x)(x)
x2 + x2=
1
2
The limit lim(x ,y)→(0,0)
u(x , y) does not exist.
Examples
lim(x ,y)→(0,0)
xy
x2 + y 2does not exist
Proof.
If (x , y) approaches (0, 0) along the x-axis, then
lim(x ,0)→(0,0)
u(x , 0) = limx→0
(x)(0)
x2 + 02= 0
However, if (x , y) approaches (0, 0) along the liney = x , then
lim(x ,x)→(0,0)
u(x , x) = limx→0
(x)(x)
x2 + x2=
1
2
The limit lim(x ,y)→(0,0)
u(x , y) does not exist.
Examples
lim(x ,y)→(0,0)
xy
x2 + y 2does not exist
Proof.
If (x , y) approaches (0, 0) along the x-axis, then
lim(x ,0)→(0,0)
u(x , 0) = limx→0
(x)(0)
x2 + 02
= 0
However, if (x , y) approaches (0, 0) along the liney = x , then
lim(x ,x)→(0,0)
u(x , x) = limx→0
(x)(x)
x2 + x2=
1
2
The limit lim(x ,y)→(0,0)
u(x , y) does not exist.
Examples
lim(x ,y)→(0,0)
xy
x2 + y 2does not exist
Proof.
If (x , y) approaches (0, 0) along the x-axis, then
lim(x ,0)→(0,0)
u(x , 0) = limx→0
(x)(0)
x2 + 02= 0
However, if (x , y) approaches (0, 0) along the liney = x , then
lim(x ,x)→(0,0)
u(x , x) = limx→0
(x)(x)
x2 + x2=
1
2
The limit lim(x ,y)→(0,0)
u(x , y) does not exist.
Examples
lim(x ,y)→(0,0)
xy
x2 + y 2does not exist
Proof.
If (x , y) approaches (0, 0) along the x-axis, then
lim(x ,0)→(0,0)
u(x , 0) = limx→0
(x)(0)
x2 + 02= 0
However, if (x , y) approaches (0, 0) along the liney = x , then
lim(x ,x)→(0,0)
u(x , x) = limx→0
(x)(x)
x2 + x2=
1
2
The limit lim(x ,y)→(0,0)
u(x , y) does not exist.
Examples
lim(x ,y)→(0,0)
xy
x2 + y 2does not exist
Proof.
If (x , y) approaches (0, 0) along the x-axis, then
lim(x ,0)→(0,0)
u(x , 0) = limx→0
(x)(0)
x2 + 02= 0
However, if (x , y) approaches (0, 0) along the liney = x , then
lim(x ,x)→(0,0)
u(x , x) = limx→0
(x)(x)
x2 + x2
=1
2
The limit lim(x ,y)→(0,0)
u(x , y) does not exist.
Examples
lim(x ,y)→(0,0)
xy
x2 + y 2does not exist
Proof.
If (x , y) approaches (0, 0) along the x-axis, then
lim(x ,0)→(0,0)
u(x , 0) = limx→0
(x)(0)
x2 + 02= 0
However, if (x , y) approaches (0, 0) along the liney = x , then
lim(x ,x)→(0,0)
u(x , x) = limx→0
(x)(x)
x2 + x2=
1
2
The limit lim(x ,y)→(0,0)
u(x , y) does not exist.
Examples
lim(x ,y)→(0,0)
xy
x2 + y 2does not exist
Proof.
If (x , y) approaches (0, 0) along the x-axis, then
lim(x ,0)→(0,0)
u(x , 0) = limx→0
(x)(0)
x2 + 02= 0
However, if (x , y) approaches (0, 0) along the liney = x , then
lim(x ,x)→(0,0)
u(x , x) = limx→0
(x)(x)
x2 + x2=
1
2
The limit lim(x ,y)→(0,0)
u(x , y) does not exist.
C as a metric space
To discuss analysis on C we must have a metricfunction defined on it.
Fortunately, we already have one!
d (z1, z2) ≡ |z1 − z2|
The open ball plays the role of the open interval!
C as a metric space
To discuss analysis on C we must have a metricfunction defined on it.
Fortunately, we already have one!
d (z1, z2) ≡ |z1 − z2|
The open ball plays the role of the open interval!
C as a metric space
To discuss analysis on C we must have a metricfunction defined on it.
Fortunately, we already have one!
d (z1, z2) ≡ |z1 − z2|
The open ball plays the role of the open interval!
C as a metric space
To discuss analysis on C we must have a metricfunction defined on it.
Fortunately, we already have one!
d (z1, z2) ≡ |z1 − z2|
The open ball plays the role of the open interval!
Continuity on the complex plane
A function f : D ⊂ C→ C is continuous atz = z0 ∈ D if ∀ε > 0, ∃δ > 0 :
|z − z0| < δ =⇒ |f (z)− f (z0)| < ε
Geometrically, this means that given any open ε-ballcentered at f (z0), Dε (f (z0)), there exists an openδ-ball centered at z0, Dδ (z0), every point of whichis carried by the map f inside Dε (f (z0)) :
f (Dδ (z0)) ⊂ Dε (f (z0))
Continuity on the complex plane
A function f : D ⊂ C→ C is continuous atz = z0 ∈ D if ∀ε > 0, ∃δ > 0 :
|z − z0| < δ =⇒ |f (z)− f (z0)| < ε
Geometrically, this means that given any open ε-ballcentered at f (z0), Dε (f (z0)), there exists an openδ-ball centered at z0, Dδ (z0), every point of whichis carried by the map f inside Dε (f (z0)) :
f (Dδ (z0)) ⊂ Dε (f (z0))
Continuity on the complex plane
x
y
z0
f
u
v
f (z0)
f maps z = x + iy to w = u + iv .
Continuity on the complex plane
x
y
z0
f
u
v
f (z0)
Given D1 = Dε (f (z0)),
Continuity on the complex plane
x
y
z0
f
u
v
f (z0)
Given D1 = Dε (f (z0)), we can find an open discDδ (z0) that is
Continuity on the complex plane
x
y
z0
f
u
v
f (z0)
Given D1 = Dε (f (z0)), we can find an open discDδ (z0) that is sufficiently small so that it maps en-tirely inside D1.
Properties of continuity
Suppose that f and g are continuous at the pointz0.
Then the following functions are continuous atz0 :
I Their sum f (z) + g(z).
I Their difference f (z)− g(z).
I Their product f (z)g(z).
I Their quotient f (z)g(z) provided that g (z0) 6= 0.
I Their composition f (g(z)) provided that f (z) iscontinuous in a neighborhood of the pointg (z0).
Properties of continuity
Suppose that f and g are continuous at the pointz0. Then the following functions are continuous atz0 :
I Their sum f (z) + g(z).
I Their difference f (z)− g(z).
I Their product f (z)g(z).
I Their quotient f (z)g(z) provided that g (z0) 6= 0.
I Their composition f (g(z)) provided that f (z) iscontinuous in a neighborhood of the pointg (z0).
Properties of continuity
Suppose that f and g are continuous at the pointz0. Then the following functions are continuous atz0 :
I Their sum f (z) + g(z).
I Their difference f (z)− g(z).
I Their product f (z)g(z).
I Their quotient f (z)g(z) provided that g (z0) 6= 0.
I Their composition f (g(z)) provided that f (z) iscontinuous in a neighborhood of the pointg (z0).
Properties of continuity
Suppose that f and g are continuous at the pointz0. Then the following functions are continuous atz0 :
I Their sum f (z) + g(z).
I Their difference f (z)− g(z).
I Their product f (z)g(z).
I Their quotient f (z)g(z) provided that g (z0) 6= 0.
I Their composition f (g(z)) provided that f (z) iscontinuous in a neighborhood of the pointg (z0).
Properties of continuity
Suppose that f and g are continuous at the pointz0. Then the following functions are continuous atz0 :
I Their sum f (z) + g(z).
I Their difference f (z)− g(z).
I Their product f (z)g(z).
I Their quotient f (z)g(z) provided that g (z0) 6= 0.
I Their composition f (g(z)) provided that f (z) iscontinuous in a neighborhood of the pointg (z0).
Properties of continuity
Suppose that f and g are continuous at the pointz0. Then the following functions are continuous atz0 :
I Their sum f (z) + g(z).
I Their difference f (z)− g(z).
I Their product f (z)g(z).
I Their quotient f (z)g(z) provided that g (z0) 6= 0.
I Their composition f (g(z)) provided that f (z) iscontinuous in a neighborhood of the pointg (z0).
Limits of complex functions
A function f : D ⊂ C→ C has a limit w0 atz = z0 ∈ D if ∀ε > 0, ∃δ > 0 :
0 < |z − z0| < δ =⇒ |f (z)− w0| < ε
We writew0 = lim
z→z0
f (z)
Corollary : A function continuous at z0 has a limitat z0 and the value of the limit is f (z0).
Limits of complex functions
A function f : D ⊂ C→ C has a limit w0 atz = z0 ∈ D if ∀ε > 0, ∃δ > 0 :
0 < |z − z0| < δ =⇒ |f (z)− w0| < ε
We writew0 = lim
z→z0
f (z)
Corollary : A function continuous at z0 has a limitat z0 and the value of the limit is f (z0).
Theorem
Let f (z) = u(x , y) + iv(x , y) be a complex functionthat is defined in some neighbourhood of z0, exceptperhaps at z0 = x0 + iy0. Then
limz→z0
f (z) = w0 = u0 + iv0
iff
lim(x ,y)→(x0,y0)
u(x , y) = u0, lim(x ,y)→(x0,y0)
v(x , y) = v0
lim f =⇒ lim u, lim v
Proof.
From the definition, ∀ε > 0, ∃δ > 0 :
0 < |z − z0| < δ =⇒ |f (z)− w0| < ε
But, f (z)− w0 = u(x , y)− u0 + i (v(x , y)− v0) sothat
|u(x , y)− u0| , |v(x , y)− v0| < |f (z)− w0|
Thus, 0 < |z − z0| < δ =⇒
|u(x , y)− u0| < ε, |v(x , y)− v0| < ε
lim f =⇒ lim u, lim v
Proof.
From the definition, ∀ε > 0, ∃δ > 0 :
0 < |z − z0| < δ =⇒ |f (z)− w0| < ε
But, f (z)− w0 = u(x , y)− u0 + i (v(x , y)− v0) sothat
|u(x , y)− u0| , |v(x , y)− v0| < |f (z)− w0|
Thus, 0 < |z − z0| < δ =⇒
|u(x , y)− u0| < ε, |v(x , y)− v0| < ε
lim f =⇒ lim u, lim v
Proof.
From the definition, ∀ε > 0, ∃δ > 0 :
0 < |z − z0| < δ =⇒ |f (z)− w0| < ε
But, f (z)− w0 = u(x , y)− u0 + i (v(x , y)− v0) sothat
|u(x , y)− u0| , |v(x , y)− v0| < |f (z)− w0|
Thus, 0 < |z − z0| < δ =⇒
|u(x , y)− u0| < ε, |v(x , y)− v0| < ε
lim f =⇒ lim u, lim v
Proof.
From the definition, ∀ε > 0, ∃δ > 0 :
0 < |z − z0| < δ =⇒ |f (z)− w0| < ε
But, f (z)− w0 = u(x , y)− u0 + i (v(x , y)− v0) sothat
|u(x , y)− u0| , |v(x , y)− v0| < |f (z)− w0|
Thus, 0 < |z − z0| < δ =⇒
|u(x , y)− u0| < ε, |v(x , y)− v0| < ε
lim u, lim v =⇒ lim f
Proof.
∀ε > 0, ∃δ1, δ2 > 0 :
0 < |z − z0| < δ1 =⇒ |u(x , y)− u0| <ε
2
0 < |z − z0| < δ2 =⇒ |v(x , y)− v0| <ε
2
Chose δ = min {δ1, δ2}.Then 0 < |z − z0| < δ =⇒
|f (z)− w0| ≤ |u(x , y)− u0|+ |v(x , y)− v0|<
ε
2+ε
2= ε
lim u, lim v =⇒ lim f
Proof.
∀ε > 0, ∃δ1, δ2 > 0 :
0 < |z − z0| < δ1 =⇒ |u(x , y)− u0| <ε
2
0 < |z − z0| < δ2 =⇒ |v(x , y)− v0| <ε
2
Chose δ = min {δ1, δ2}.Then 0 < |z − z0| < δ =⇒
|f (z)− w0| ≤ |u(x , y)− u0|+ |v(x , y)− v0|<
ε
2+ε
2= ε
lim u, lim v =⇒ lim f
Proof.
∀ε > 0, ∃δ1, δ2 > 0 :
0 < |z − z0| < δ1 =⇒ |u(x , y)− u0| <ε
2
0 < |z − z0| < δ2 =⇒ |v(x , y)− v0| <ε
2
Chose δ = min {δ1, δ2}.
Then 0 < |z − z0| < δ =⇒
|f (z)− w0| ≤ |u(x , y)− u0|+ |v(x , y)− v0|<
ε
2+ε
2= ε
lim u, lim v =⇒ lim f
Proof.
∀ε > 0, ∃δ1, δ2 > 0 :
0 < |z − z0| < δ1 =⇒ |u(x , y)− u0| <ε
2
0 < |z − z0| < δ2 =⇒ |v(x , y)− v0| <ε
2
Chose δ = min {δ1, δ2}.Then 0 < |z − z0| < δ =⇒
|f (z)− w0| ≤ |u(x , y)− u0|+ |v(x , y)− v0|
<ε
2+ε
2= ε
lim u, lim v =⇒ lim f
Proof.
∀ε > 0, ∃δ1, δ2 > 0 :
0 < |z − z0| < δ1 =⇒ |u(x , y)− u0| <ε
2
0 < |z − z0| < δ2 =⇒ |v(x , y)− v0| <ε
2
Chose δ = min {δ1, δ2}.Then 0 < |z − z0| < δ =⇒
|f (z)− w0| ≤ |u(x , y)− u0|+ |v(x , y)− v0|<
ε
2+ε
2
= ε
lim u, lim v =⇒ lim f
Proof.
∀ε > 0, ∃δ1, δ2 > 0 :
0 < |z − z0| < δ1 =⇒ |u(x , y)− u0| <ε
2
0 < |z − z0| < δ2 =⇒ |v(x , y)− v0| <ε
2
Chose δ = min {δ1, δ2}.Then 0 < |z − z0| < δ =⇒
|f (z)− w0| ≤ |u(x , y)− u0|+ |v(x , y)− v0|<
ε
2+ε
2= ε
Some properties of limits
Let limz→z0
f (z) = A and limz→z0
g(z) = B . Then
I limz→z0
[f (z)± g(z)] = A± B
I limz→z0
f (z)g(z) = AB
I limz→z0
f (z)
g(z)=
A
B, where B 6= 0.
Some properties of limits
Let limz→z0
f (z) = A and limz→z0
g(z) = B . Then
I limz→z0
[f (z)± g(z)] = A± B
I limz→z0
f (z)g(z) = AB
I limz→z0
f (z)
g(z)=
A
B, where B 6= 0.
Some properties of limits
Let limz→z0
f (z) = A and limz→z0
g(z) = B . Then
I limz→z0
[f (z)± g(z)] = A± B
I limz→z0
f (z)g(z) = AB
I limz→z0
f (z)
g(z)=
A
B, where B 6= 0.
Some properties of limits
Let limz→z0
f (z) = A and limz→z0
g(z) = B . Then
I limz→z0
[f (z)± g(z)] = A± B
I limz→z0
f (z)g(z) = AB
I limz→z0
f (z)
g(z)=
A
B, where B 6= 0.
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