Christopher Croke
University of Pennsylvania
Limits and Continuity/Partial Derivatives
Math 115
Calculus 115
Limits
For (x0 , y0 ) an interior or a boundary point of the domain of a
function f (x, y ).
Definition:
lim
f (x, y ) = L
(x,y )→(x0 ,y0 )
if for every > 0 there is a δ > 0 such that: for all (x, y ) in the
domain of f if
q
0 < (x − x0 )2 + (y − y0 )2 < δ
then
|f (x, y ) − L| < .
Calculus 115
Limits
For (x0 , y0 ) an interior or a boundary point of the domain of a
function f (x, y ).
Definition:
lim
f (x, y ) = L
(x,y )→(x0 ,y0 )
if for every > 0 there is a δ > 0 such that: for all (x, y ) in the
domain of f if
q
0 < (x − x0 )2 + (y − y0 )2 < δ
then
|f (x, y ) − L| < .
Calculus 115
This definition is really the same as in one dimension and so
satisfies the same rules with respect to +, −, ×, ÷. See for
example page 803 for a list.
Calculus 115
This definition is really the same as in one dimension and so
satisfies the same rules with respect to +, −, ×, ÷. See for
example page 803 for a list.
For example:
if
lim
f (x, y ) = L and
lim
g (x, y ) = K
(x,y )→(x0 ,y0 )
(x,y )→(x0 ,y0 )
Calculus 115
This definition is really the same as in one dimension and so
satisfies the same rules with respect to +, −, ×, ÷. See for
example page 803 for a list.
For example:
if
lim
f (x, y ) = L and
lim
g (x, y ) = K
(x,y )→(x0 ,y0 )
(x,y )→(x0 ,y0 )
then
lim
f (x, y ) + g (x, y ) = L + K
(x,y )→(x0 ,y0 )
x 2 + 2y + 3
=?
x +y
(x,y )→(1,0)
lim
Calculus 115
This definition is really the same as in one dimension and so
satisfies the same rules with respect to +, −, ×, ÷. See for
example page 803 for a list.
For example:
if
lim
f (x, y ) = L and
lim
g (x, y ) = K
(x,y )→(x0 ,y0 )
(x,y )→(x0 ,y0 )
then
lim
f (x, y ) + g (x, y ) = L + K
(x,y )→(x0 ,y0 )
x 2 + 2y + 3
=?
x +y
(x,y )→(1,0)
lim
x2 − y2
=?
(x,y )→(0,0) x − y
lim
Calculus 115
Continuity(The definition is really the same.)
f is continuous at (x0 , y0 ) if
lim
(x,y )→(x0 ,y0 )
f (x, y ) = f (x0 , y0 ).
This means three things:
Calculus 115
Continuity(The definition is really the same.)
f is continuous at (x0 , y0 ) if
lim
(x,y )→(x0 ,y0 )
f (x, y ) = f (x0 , y0 ).
This means three things:
f is defined at (x0 , y0 ).
Calculus 115
Continuity(The definition is really the same.)
f is continuous at (x0 , y0 ) if
lim
(x,y )→(x0 ,y0 )
f (x, y ) = f (x0 , y0 ).
This means three things:
f is defined at (x0 , y0 ).
lim(x,y )→(x0 ,y0 ) f (x, y ) exists.
Calculus 115
Continuity(The definition is really the same.)
f is continuous at (x0 , y0 ) if
lim
(x,y )→(x0 ,y0 )
f (x, y ) = f (x0 , y0 ).
This means three things:
f is defined at (x0 , y0 ).
lim(x,y )→(x0 ,y0 ) f (x, y ) exists.
They are equal.
Calculus 115
Continuity(The definition is really the same.)
f is continuous at (x0 , y0 ) if
lim
(x,y )→(x0 ,y0 )
f (x, y ) = f (x0 , y0 ).
This means three things:
f is defined at (x0 , y0 ).
lim(x,y )→(x0 ,y0 ) f (x, y ) exists.
They are equal.
There is an added complication here. Consider
y
if (x, y ) 6= (0, 0)
x 2 +y 2
f (x, y ) =
0
if (x, y ) = (0, 0)
Calculus 115
If f (x, y ) has different limits along two different paths in the
domain of f as (x, y ) approaches (x0 , y0 ) then
lim(x,y )→(x0 ,y0 ) f (x, y ) does not exist.
Calculus 115
If f (x, y ) has different limits along two different paths in the
domain of f as (x, y ) approaches (x0 , y0 ) then
lim(x,y )→(x0 ,y0 ) f (x, y ) does not exist.
It is not enough to check only along straight lines! See the
example in the text.
Calculus 115
If f (x, y ) has different limits along two different paths in the
domain of f as (x, y ) approaches (x0 , y0 ) then
lim(x,y )→(x0 ,y0 ) f (x, y ) does not exist.
It is not enough to check only along straight lines! See the
example in the text.
Continuity acts nicely under compositions:
If f is continuous at (x0 , y0 ) and g (a function of a single variable)
is continuous at f (x0 , y0 ) then g ◦ f is continuous at (x0 , y0 ).
Calculus 115
If f (x, y ) has different limits along two different paths in the
domain of f as (x, y ) approaches (x0 , y0 ) then
lim(x,y )→(x0 ,y0 ) f (x, y ) does not exist.
It is not enough to check only along straight lines! See the
example in the text.
Continuity acts nicely under compositions:
If f is continuous at (x0 , y0 ) and g (a function of a single variable)
is continuous at f (x0 , y0 ) then g ◦ f is continuous at (x0 , y0 ).
Example:
y2
sin( 2
)
x −1
Calculus 115
Partial Derivatives of f (x, y )
∂f
∂x
”partial derivative of f with respect to x”
Calculus 115
Partial Derivatives of f (x, y )
∂f
∂x
”partial derivative of f with respect to x”
Easy to calculate: just take the derivative of f w.r.t. x thinking of
y as a constant.
∂f
∂y
”partial derivative of f with respect to y ”
Calculus 115
Examples: Compute
∂f
∂x
and
∂f
∂y
for f (x, y ) =
2x 2 + 4xy
Calculus 115
Examples: Compute
∂f
∂x
and
∂f
∂y
for f (x, y ) =
2x 2 + 4xy
(xy 3 + xy )3
Calculus 115
Examples: Compute
∂f
∂x
and
∂f
∂y
for f (x, y ) =
2x 2 + 4xy
(xy 3 + xy )3
sin(xy + y )
Calculus 115
Examples: Compute
∂f
∂x
and
∂f
∂y
for f (x, y ) =
2x 2 + 4xy
(xy 3 + xy )3
sin(xy + y )
x 3y
y + x2
Calculus 115
Example: For f (x, y ) = 5 + 2xy − 3x 2 y 2 find
∂f
(1, 2)
∂x
Calculus 115
Example: For f (x, y ) = 5 + 2xy − 3x 2 y 2 find
∂f
(1, 2)
∂x
∂f
(2, 0)
∂y
Calculus 115
Example: For f (x, y ) = 5 + 2xy − 3x 2 y 2 find
∂f
(1, 2)
∂x
∂f
(2, 0)
∂y
This notion shows up in many fields and many notations have been
developed for the same thing.
∂f
∂x
fx
zx
∂z
∂x
Calculus 115
∂x f
What does
∂f
∂x (a, b)
mean geometrically?
Calculus 115
What does
∂f
∂x (a, b)
mean geometrically?
Calculus 115
Definition:
d
∂f
(a, b) =
|x=a f (x, b) =
∂x
dx
Calculus 115
Definition:
d
∂f
(a, b) =
|x=a f (x, b) =
∂x
dx
f (a + h, b) − f (a, b)
h→0
h
= lim
Calculus 115
Definition:
d
∂f
(a, b) =
|x=a f (x, b) =
∂x
dx
f (a + h, b) − f (a, b)
h→0
h
= lim
Calculus 115
Similarly
∂f
d
f (a, b + h) − f (a, b)
(a, b) =
|y =b f (a, y ) = lim
h→0
∂y
dy
h
Calculus 115
Similarly
∂f
d
f (a, b + h) − f (a, b)
(a, b) =
|y =b f (a, y ) = lim
h→0
∂y
dy
h
Calculus 115
Calculus 115
You can do implicit differentiation in the same way as for functions
of one variable.
Calculus 115
You can do implicit differentiation in the same way as for functions
of one variable.
∂z
when z is is defined implicitly as a function of x
Example: Find ∂x
and y by the formula;
x 2 + y 2 + z 2 = 5.
Calculus 115
For f (x, y ) since fx and fy are functions of x and y we can take
partial derivatives of these to yield second partial derivatives.
Calculus 115
For f (x, y ) since fx and fy are functions of x and y we can take
partial derivatives of these to yield second partial derivatives.
Notation
∂ ∂f
∂2f
)=
.
∂x ∂x
∂x∂x
Calculus 115
For f (x, y ) since fx and fy are functions of x and y we can take
partial derivatives of these to yield second partial derivatives.
Notation
∂ ∂f
∂2f
)=
.
∂x ∂x
∂x∂x
∂ ∂f
∂2f
)=
.
∂y ∂x
∂y ∂x
Calculus 115
For f (x, y ) since fx and fy are functions of x and y we can take
partial derivatives of these to yield second partial derivatives.
Notation
∂ ∂f
∂2f
)=
.
∂x ∂x
∂x∂x
∂ ∂f
∂2f
)=
.
∂y ∂x
∂y ∂x
∂ ∂f
∂2f
)=
.
∂x ∂y
∂x∂y
Calculus 115
For f (x, y ) since fx and fy are functions of x and y we can take
partial derivatives of these to yield second partial derivatives.
Notation
∂ ∂f
∂2f
)=
.
∂x ∂x
∂x∂x
∂ ∂f
∂2f
)=
.
∂y ∂x
∂y ∂x
∂ ∂f
∂2f
)=
.
∂x ∂y
∂x∂y
∂ ∂f
∂2f
)=
.
∂y ∂y
∂y ∂y
Calculus 115
Example: Compute all second partials of
f (x, y ) = x 3 + 2xy 2 − y 2 + 1.
Calculus 115
Example: Compute all second partials of
f (x, y ) = x 3 + 2xy 2 − y 2 + 1.
Theorem
Mixed derivative theorem If f (x, y ),fx ,fy ,fxy ,and fyx are defined in
an open region about (a, b) and all are continuous at (a, b) then
fxy (a, b) = fyx (a, b)
Calculus 115
Example: Compute all second partials of
f (x, y ) = x 3 + 2xy 2 − y 2 + 1.
Theorem
Mixed derivative theorem If f (x, y ),fx ,fy ,fxy ,and fyx are defined in
an open region about (a, b) and all are continuous at (a, b) then
fxy (a, b) = fyx (a, b)
Moral: ”Can differentiate in any order”
Calculus 115
Example: Compute all second partials of
f (x, y ) = x 3 + 2xy 2 − y 2 + 1.
Theorem
Mixed derivative theorem If f (x, y ),fx ,fy ,fxy ,and fyx are defined in
an open region about (a, b) and all are continuous at (a, b) then
fxy (a, b) = fyx (a, b)
Moral: ”Can differentiate in any order”
This also holds for more derivatives:
fxyxx
∂3f
∂x∂y 2
Calculus 115
This all works for functions with more variables.
Calculus 115
This all works for functions with more variables. Example:
Compute fxyz for f = xyz + e xz xsin(z) + x 3 ln(z).
Calculus 115
This all works for functions with more variables. Example:
Compute fxyz for f = xyz + e xz xsin(z) + x 3 ln(z).
It turns out that just because fx (a, b) and fy (a, b) exist does not
mean that f is continuous at (a, b):
Calculus 115
This all works for functions with more variables. Example:
Compute fxyz for f = xyz + e xz xsin(z) + x 3 ln(z).
It turns out that just because fx (a, b) and fy (a, b) exist does not
mean that f is continuous at (a, b):
1 if xy = 0
f (x, y ) =
0 if xy 6= 0
Calculus 115
This all works for functions with more variables. Example:
Compute fxyz for f = xyz + e xz xsin(z) + x 3 ln(z).
It turns out that just because fx (a, b) and fy (a, b) exist does not
mean that f is continuous at (a, b):
1 if xy = 0
f (x, y ) =
0 if xy 6= 0
It turns out that one has to b more careful (see text) about what it
means to be differentiable at (a, b) but it is enough if fx and fy
exist and are continuous in a disk about (a, b).
Calculus 115
This all works for functions with more variables. Example:
Compute fxyz for f = xyz + e xz xsin(z) + x 3 ln(z).
It turns out that just because fx (a, b) and fy (a, b) exist does not
mean that f is continuous at (a, b):
1 if xy = 0
f (x, y ) =
0 if xy 6= 0
It turns out that one has to b more careful (see text) about what it
means to be differentiable at (a, b) but it is enough if fx and fy
exist and are continuous in a disk about (a, b).
If f is differentiable at (a, b) then it will be continuous at (a, b).
Calculus 115