Limits and Continuity/Partial Derivatives

Christopher Croke

University of Pennsylvania

Math 115 UPenn, Fall 2011

Christopher Croke 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 2 2 0 < (x − x0) + (y − y0) < δ

then |f (x, y) − L| < .

Christopher Croke Calculus 115 Limits

then |f (x, y) − L| < .

Christopher Croke Calculus 115 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)

x2 + 2y + 3 lim =? (x,y)→(1,0) x + y

x2 − y 2 lim =? (x,y)→(0,0) x − y

This deﬁnition is really the same as in one dimension and so satisﬁes the same rules with respect to +, −, ×, ÷. See for example page 775 for a list.

Christopher Croke Calculus 115 then lim f (x, y) + g(x, y) = L + K (x,y)→(x0,y0)

x2 + 2y + 3 lim =? (x,y)→(1,0) x + y

x2 − y 2 lim =? (x,y)→(0,0) x − y

This deﬁnition is really the same as in one dimension and so satisﬁes the same rules with respect to +, −, ×, ÷. See for example page 775 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)

Christopher Croke Calculus 115 x2 − y 2 lim =? (x,y)→(0,0) x − y

x2 + 2y + 3 lim =? (x,y)→(1,0) x + y

Christopher Croke 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 775 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)

x2 + 2y + 3 lim =? (x,y)→(1,0) x + y

x2 − y 2 lim =? (x,y)→(0,0) x − y

Christopher Croke Calculus 115 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) f (x, y) = x2+y 2 0 if (x, y) = (0, 0)

Continuity(The deﬁnition is really the same.)

f is continuous at (x0, y0) if

lim f (x, y) = f (x0, y0). (x,y)→(x0,y0)

This means three things:

Christopher Croke Calculus 115 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) f (x, y) = x2+y 2 0 if (x, y) = (0, 0)

Continuity(The deﬁnition is really the same.)

f is continuous at (x0, y0) if

lim f (x, y) = f (x0, y0). (x,y)→(x0,y0)

This means three things:

f is deﬁned at (x0, y0).

Christopher Croke Calculus 115 They are equal.

There is an added complication here. Consider

y if (x, y) 6= (0, 0) f (x, y) = x2+y 2 0 if (x, y) = (0, 0)

Continuity(The deﬁnition is really the same.)

f is continuous at (x0, y0) if

lim f (x, y) = f (x0, y0). (x,y)→(x0,y0)

This means three things:

f is deﬁned at (x0, y0).

lim(x,y)→(x0,y0) f (x, y) exists.

Christopher Croke Calculus 115 There is an added complication here. Consider

y if (x, y) 6= (0, 0) f (x, y) = x2+y 2 0 if (x, y) = (0, 0)

Continuity(The deﬁnition is really the same.)

f is continuous at (x0, y0) if

lim f (x, y) = f (x0, y0). (x,y)→(x0,y0)

This means three things:

f is deﬁned at (x0, y0).

lim(x,y)→(x0,y0) f (x, y) exists. They are equal.

Christopher Croke Calculus 115 Continuity(The deﬁnition is really the same.)

f is continuous at (x0, y0) if

lim f (x, y) = f (x0, y0). (x,y)→(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) f (x, y) = x2+y 2 0 if (x, y) = (0, 0)

Christopher Croke Calculus 115 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: y 2 sin( ) x2 − 1

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.

Christopher Croke Calculus 115 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: y 2 sin( ) x2 − 1

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.

Christopher Croke Calculus 115 Example: y 2 sin( ) x2 − 1

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).

Christopher Croke 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.

Christopher Croke Calculus 115 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”

Partial Derivatives of f (x, y)

∂f ∂x ”partial derivative of f with respect to x”

Christopher Croke 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”

Christopher Croke Calculus 115 (xy 3 + xy)3

sin(xy + y)

x3y y + x2

∂f ∂f Examples: Compute ∂x and ∂y for f (x, y) =

2x2 + 4xy

Christopher Croke Calculus 115 sin(xy + y)

x3y y + x2

∂f ∂f Examples: Compute ∂x and ∂y for f (x, y) =

2x2 + 4xy

(xy 3 + xy)3

Christopher Croke Calculus 115 x3y y + x2

∂f ∂f Examples: Compute ∂x and ∂y for f (x, y) =

2x2 + 4xy

(xy 3 + xy)3

sin(xy + y)

Christopher Croke Calculus 115 ∂f ∂f Examples: Compute ∂x and ∂y for f (x, y) =

2x2 + 4xy

(xy 3 + xy)3

sin(xy + y)

x3y y + x2

Christopher Croke Calculus 115 ∂f (2, 0) ∂y

This notion shows up in many ﬁelds and many notations have been developed for the same thing.

∂f ∂z f z ∂ f ∂x x x ∂x x

Example: For f (x, y) = 5 + 2xy − 3x2y 2 ﬁnd

∂f (1, 2) ∂x

Christopher Croke Calculus 115 This notion shows up in many ﬁelds and many notations have been developed for the same thing.

∂f ∂z f z ∂ f ∂x x x ∂x x

Example: For f (x, y) = 5 + 2xy − 3x2y 2 ﬁnd

∂f (1, 2) ∂x

∂f (2, 0) ∂y

Christopher Croke Calculus 115 Example: For f (x, y) = 5 + 2xy − 3x2y 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 ∂z f z ∂ f ∂x x x ∂x x

Christopher Croke Calculus 115 ∂f What does ∂x (a, b) mean geometrically?

Christopher Croke Calculus 115 f (a + h, b) − f (a, b) = lim h→0 h Similarly

∂f d f (a, b + h) − f (a, b) (a, b) = |y=bf (a, y) = lim ∂y dy h→0 h

Deﬁnition: ∂f d (a, b) = | f (x, b) = ∂x dx x=a

Christopher Croke Calculus 115 Similarly

∂f d f (a, b + h) − f (a, b) (a, b) = |y=bf (a, y) = lim ∂y dy h→0 h

Deﬁnition: ∂f d (a, b) = | f (x, b) = ∂x dx x=a f (a + h, b) − f (a, b) = lim h→0 h

Christopher Croke Calculus 115 Deﬁnition: ∂f d (a, b) = | f (x, b) = ∂x dx x=a f (a + h, b) − f (a, b) = lim h→0 h Similarly

∂f d f (a, b + h) − f (a, b) (a, b) = |y=bf (a, y) = lim ∂y dy h→0 h

Christopher Croke Calculus 115 ∂z Example: Find ∂x when z is is deﬁned implicitly as a function of x and y by the formula;

x2 + y 2 + z2 = 5.

You can do implicit diﬀerentiation in the same way as for functions of one variable.

Christopher Croke Calculus 115 You can do implicit diﬀerentiation in the same way as for functions of one variable. ∂z Example: Find ∂x when z is is deﬁned implicitly as a function of x and y by the formula;

x2 + y 2 + z2 = 5.

Christopher Croke Calculus 115 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

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.

Christopher Croke Calculus 115 ∂ ∂f ∂2f ) = . ∂y ∂x ∂y∂x

∂ ∂f ∂2f ) = . ∂x ∂y ∂x∂y

∂ ∂f ∂2f ) = . ∂y ∂y ∂y∂y

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

Christopher Croke Calculus 115 ∂ ∂f ∂2f ) = . ∂x ∂y ∂x∂y

∂ ∂f ∂2f ) = . ∂y ∂y ∂y∂y

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

Christopher Croke Calculus 115 ∂ ∂f ∂2f ) = . ∂y ∂y ∂y∂y

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

Christopher Croke 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

Christopher Croke Calculus 115 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

Example: Compute all second partials of f (x, y) = x3 + 2xy 2 − y 2 + 1.

Christopher Croke Calculus 115 Moral: ”Can diﬀerentiate in any order” This also holds for more derivatives:

fxyxx ∂3f ∂x∂y 2

Example: Compute all second partials of f (x, y) = x3 + 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)

Christopher Croke Calculus 115 This also holds for more derivatives:

fxyxx ∂3f ∂x∂y 2

Example: Compute all second partials of f (x, y) = x3 + 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”

Christopher Croke Calculus 115 Example: Compute all second partials of f (x, y) = x3 + 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

Christopher Croke Calculus 115 Example: xz 3 Compute fxyz for f = xyz + e xsin(z) + x 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).

This all works for functions with more variables.

Christopher Croke Calculus 115 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

This all works for functions with more variables. Example: xz 3 Compute fxyz for f = xyz + e xsin(z) + x ln(z).

Christopher Croke Calculus 115 1 if xy = 0 f (x, y) = 0 if xy 6= 0

This all works for functions with more variables. Example: xz 3 Compute fxyz for f = xyz + e xsin(z) + x 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):

Christopher Croke Calculus 115 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).

This all works for functions with more variables. Example: xz 3 Compute fxyz for f = xyz + e xsin(z) + x 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

Christopher Croke Calculus 115 If f is diﬀerentiable at (a, b) then it will be continuous at (a, b).

This all works for functions with more variables. Example: xz 3 Compute fxyz for f = xyz + e xsin(z) + x 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).

Christopher Croke Calculus 115 This all works for functions with more variables. Example: xz 3 Compute fxyz for f = xyz + e xsin(z) + x 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

Christopher Croke Calculus 115