Chapter 11
Three-dimensional
Analytic Geometry and Vectors
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11.5 Partial derivatives and total differential
1. Let f (x, y) be a function of two variables, and suppose
that x vary while keeping y fixed, say y=b, where b is a
constant. Then we have a function of single variable x,
namely g(x)= f (x, b). If g(x) has a derivative at a, then we
call the derivative the partial derivative of f with respect to
x at (a, b) and denoted it by fx(a, b). Thus we have
fx(a,b)=g'(a) where g(x)= f (x, b) or
f (a  h, b)  f (a, b)
f x (a, b)  lim
h 0
h
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similarly,
f ( a, b  h)  f ( a, b)
f y (a, b)  lim
h
h 0
is called partial derivative of f(x, y)with respect y
at (a, b).
2. If let the point (a,b) vary, fx and fy become
functions of two variables. So we can get the
following:
If fx is a function of two variables, its partial
derivatives are functions fx and fy defined by
f ( x  h, y)  f ( x, y)
fx(x,y)= lim
h0
h
fy(x,y)= lim f ( x, y  h)  f ( x, y)
h0
h
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There are other notations for partial derivatives:
If z=f(x, y),
f 
z
fx(x,y)= fx= f1=D1f =Dxf =
 f ( x, y) 
x x
x
f 
z

f
(
x
,
y
)

fy(x,y)= fy=f2 = D2f =Dyf = y y
y
Rule for finding partial derivatives of z=f(x,y):
<1>. To find fx , regard y as a constant and
differentiate f(x, y) with respect with x.
<2>. To find fy , regard x as a constant and
differentiate f(x, y) with respect with y.
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Example If f(x, y)=xsin(x2+y2)exy, find fx(2, 1) and fy(1,
2).
Example If f(x, y)=cos( x ), find fx(x, y) and fy(x, y).
1 y
Similarly, we can define partial derivatives of function of
more than two variables. If u=f(x1, x2 , ..., xn), then
f ( x1, x2 ,..., xi1, xi  h, xi1,...xn )  f ( x1, x2 ,..., xi1, xi , xi1,...xn )
u
fi  f x   lim
xi h0
h
i
i=1,2,....,n
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3.We can also define fxx=(fx )x , fxy=(fx )y , and so on.
 f  2 f  2u  u
fxx= y ( x )  yx  yx  y ( x )
 f  2 f  2u  u
fxy= x ( x )  2  2  x ( x )
x x
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Definition If z=f(x, y), then f is called differential at
(a, b) if z = fx(a, b) x+ fy(a, b) y+1x+2 y
where 1 and 2 0 as (x, y)0 .
Then generally fx(x, y)x+ fy(x, y)y is called total
differential, denoted by dz or df . That is
dz=df = fx(x, y)x+ fy(x, y)y
Example If z=f(x, y)=2x2+y2, find total differential dz.
Similarly, we can define total differential of function
of more than two variables.
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11.6 The Chain Rule
1. The Chain rule (case I) Suppose that z=f(x, y) is a
differentiable function of x and y, where x=g(t) and y=h(t)
are both differentiable functions of t. Then z is a
differentiable function of t and
dz f dx f dy


dt x dt y dt
Proof A change of t produces changes of x in
x and y in y. Then it produces change of z in
z,and from the definition we have
f
f
z  x  y  1x   2y
x
y
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z f x f y
x
y


 1   2
t x t y t
t
t
We let t0, then x=g(t+t) - g(t) 0，because g is differentiable
and therefore continuous. Similarly, y0. This means that 10 ,
20 , so
dz f dx f dy


dt x dt y dt
2. If z=x2y+3xy4, where x=et , y=sint , find dz/dx.
3. Suppose chain rule (case 2) Suppose that z=f(x, y)
is differentiable, x=g(t, s), y=h(t, s), and the partial
derivatives, gt, gs , ht and hs exist. Then
z f x f y


t x t y t
z f x f y


s x s y s
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4. The chain rule (General version). Suppose u=f(x1,x2 ,... ,xn),
and each xi is a function of m variables t1, t2 ,... , tm , such that
the all partial derivatives, xj ti, exist (j=1, 2,...., m). Then
u is a function of t1, t2 ,... , tm , and
u u x1 u x2
u xm


 ...... 
t j x1 t j x2 t j
xm t j
for , j =1,2,.... ,m.
5. If u=x4y+y2z3, where x =rset,y=rs2e-t , z=r2ssint , find
u s when r=2, s=1, t=0.
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6. Implicit differentiation We suppose that an
equation of the form F(x, y)=0, define y implicitly as
a differentiable function of x , that is , y=f(x), where
F(x, f(x))0 for all x in the domain of f. If is F(x, y)
is differentiable and F y0, then
F
dy
  x
dx
F
y
7. Example Find y ' if x3+y3=6xy.
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8. More generally, We suppose that z is given
implicitly by an equation of the form F(x, y, z)=0. This
means that F(x, y, f(x, y))0 for all (x, y) in the domain of
f. If F(x, y, z) is differentiable and F z0, then
F
z
y

y
F
z
F
z
  x ,
x
F
z
9. Example Find
z
x
and
z
y
if x3+y3 +z3 =6xy.
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11.7 Directional derivatives and the gradient
Definition The directional derivative of at (x0, y0) in the
direction of a unit vector u=<a, b> is
f ( x0  ha, y0  hb)  f ( x0 , y0 )
Du f ( x0 , y0 )  lim
h 0
h
if this limit exists.
Theorem If f is differentiable function of x and y, then f
has directional derivative in the direction of any vector
unit vector u=<a, b> and
Du f ( x, y)  f x ( x, y)a  f y ( x, y)b
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Example Find the directional derivative Duf(x, y) if f(x,
y)=x3-3xy+4y2 and u is the unit vector given by angle
=/6. What is Duf(1, 2) ?
Definition If f is a function of two variables x and y,
then the gradient of f is the vector function f(x, y)
defined by
f(x, y)=< fx(x, y) , fy(x, y)
f
f
i

j
>=x
y
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Example If f(x, y)=sinx+exy , find the f(x, y).
Corollary u=<a,b> , Duf(x, y) =f(x, y)·u.
Example If f(x, y)=x2 y3 –4y , find the directional
derivative of f(x, y) at (2, –1) in the direction v=2i+5j,.
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For functions of three variables we can define directional
derivatives in a similar manner.
Definition The directional derivative of at (x0, y0 , z0) in
the direction of a unit vector u=<a, b, c> is
f ( x0  ha, y0  hb, z 0  hc)  f ( x0 , y0 , z 0 )
Du f ( x0 , y0 , z 0 )  lim
h 0
h
if this limit exists. We can use vector notation as following:
P0=< x0, y0 , z0 > (or P0=< x0, y0> ) and
f ( P0  hu )  f ( P0 )
Du f ( P0 )  lim
h 0
h
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Definition If f is a function of three variables x, y and z, then
the gradient of f is the vector function, denoted by f(x, y) or
gradf, defined by
gradf =f(x, y)=< fx , fy,
f
f
f
i

j

k
fz >=
x
y
z
Corollary: u=<a, b, c> , Duf(x, y, z) =f(x, y, z)·u.
Example If f(x, y, z)=xsin(z y), find (a) the directional
derivative of f(x, y, z) at (1, 3, 0) in the direction v=1i+2j–k.
(b) the gradient of f .
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Theorem Suppose f is a differentiable function of two or
three variables. The maximum value of the directional
derivative Duf(P) is |f(P)| and it occurs when u has the
same direction as the gradient vector f(P) .
Example (a) If f(x, y)=xey, find the rate of change at p(2, 0)
in direction from P to Q(1/2, 2).
(b)In what direction does f have the maximum rate of
change ? What is this maximum rate of change. p807
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Tangent planes to level surface: (ommited)
11.8 Maximum and minimum values p812
Definition A function of two or three variables has a
local maximum at (a, b) (or (a, b, c)) if f(x, y) f(a, b)
(or f(x, y, z) f(a, b, c) ) for all (x, y) (or (x, y, z)) in
such a small disk with center (a, b) (or (a, b, c)) . If f(x,
y) f(a, b) (or f(x, y, z)  f(a, b, c) ) for all (x, y) (or (x,
y, z)) in such a small disk with center (a, b) (or (a, b,
c)) , f(a, b) (or f(a, b, c) ) is a local minimum value.
If for all points (x, y) (or (x, y, z)) in the domain of f ,
then has an absolute maximum (or absolute
minimum) at point (a, b) (or (a, b, c)) .
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Theorem If f has a local extremum (this is local minimum or
local maximum) (a,b) and the first-order partial derivatives of
f exists there, then fx(a,b)=0 and fy(a,b)=0 .
A point (a,b) such that fx(a,b)=0 and fy(a,b)=0, or one of
these partial derivatives does not exist, is called critical point
(or stationary point) .
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Example Find the extremum values of f(x,y)=y2-x2,
f(0,0)=0 can not be an extremum value for f, so f has no
extremum values. But (0,0) is a the critical point of f.
Example Find the extremum values of f(x,y)=y2+x2 2x-6y+14.
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The Second derivative test Suppose the second partial
derivatives of f are continuous in a disk with center (a, b)
and (a, b) is the critical point. Let
D=D(a,b) = fxx(a, b) fyy(a, b) –[fxy(a, b)]2
(a) If D>0 and fxx(a, b)>0, then f(a, b) is a local minimum.
(b) If D>0 and fxx(a, b)<0, then f(a, b) is a local maximum.
(c) If D<0, then f(a, b) is not extremum, then (a, b) is
called a saddle point.
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Example Find the local extremum of f(x,y)=x4+y4 -4xy+1.
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11.9 Lagrange multiplier P821
Suppose now that we want to find the maximum and
minimum values of f(x, y, z) subject to two constrain
(side conditions) of form g(x, y, z)=k and h(x, y, z)=c.
Example Find the maximum value of the function
V=xyz, subject to the constrain 2xz+2yz+xy=12.
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To find the the maximum and minimum values of f(x,y,z)
subject to two constrain g(x,y,z)=k and h(x,y,z)=c(assuming
that the extrema exist):
(a) Find all values of x, y, z, and , , such that f(x, y, z)= 
g(x, y, z)+  h(x, y, z) and g(x, y, z)=k, h(x, y, z)=c;
(b) Evaluate f at all the points (x, y, z) that arise from step
(a). The largest of these values is the maximum value of f .
The smallest of these values is the minimum value of f .
( and  are called Lagrange multiplier. This method is
called the method of Lagrange multipliers.
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Example Find the maximum value of the function
f(x, y, z)=x+2y+3z on the curve of intersection of the
plane x–y+ z=1 and the cylinder x2+y2=1. (p826)
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11.1 Three-dimensional coordinate systems
1. Recall coordinate systems in plane which
locate the points in plane by an ordered pair (a,b) of
real number, where a is the x-coordinate and b is the
y-coordinate.
2. We can use a similar idea to locate the points in
space by an ordered triple (a,b,c). We first choose a
fixed point O (called origin) and three axes(directed
lines) through O that are perpendicular to each other,
called the coordinate axes and labeled the x-axis, yaxis, and z-axis.
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The three coordinate
z
planes divide space
into eight parts,
called octants. They
are called the first,
the second,... ,the
eighth octant
xz-plane
respectively,
III
II
VI
illustrated in the
figure.
yz-plane
x
y
oxy-plane
VIII
V
I
IV
VII
x-axis and y-axis form a plane coordinate and the
direction of z-axis is determined by the Right-hand rule
For every point P in
space, there is only
an ordered triple
(a,b,c) of real
numbers by three
planes which are
parallel to three
coordinate planes
respectively
through P. a,b,c are
interceptes with
three coordinate
axes, respectively.
This is called three-dimensional rectangular
coordinate system
z
(0,0,c) - c
(a,0,c)
-1 1
+
-1
a_
(a,0,0)
(0,b,c)
P (a,b,c)
b
+
oxy-plane
y
(a,b,0)
x
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Contrary to this,
for every ordered
triple (a,b,c),
there is only one
point P, common
point of three
planes which are
perpendicular to
three coordinate
axes respectively
and through
(a,0,0), (0,b,0),
(0,0,c),
respectively.
z
o
(a,0,0) a
(0,0,c) - c
(0,b,0)
b
y
(a,b,0)
x
(a,b,c)
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2. We now can get a distance formula: Let P1(x1, y1 , z1) and
P2 (x2 , y2 , z2 ) be the points in space, Then the distance |
P1P2| between the two points is
| P1P2|= ( x1  x2 )2  ( y1  y2 )2  ( z1  z2 )2
P2(x2, y2 , z2)
Proof:
z
z2
P2
z1
P1
P1(x1, y1 , z1)
A
x1
x2
y1
B
y2
y
3. Example
Find the distance from P(2,-1,7) to Q(1,-3,5).
4. Find an equation of a sphere with radius r and center C(h,k,l).
SOLUTION: Let P(x, y, z) be a point in the sphere. So the distance
|PC|=r. Then we can easily obtain that
(x- h) 2+ (y- h) 2 + (z- h)2= r2.
5.
From 4, every point in the sphere, the coordinate of the point
must satisfies the Equation. On other hand, every ordered triple
(x, y, z) which satisfies the Equation above, the point that is
determined by (x, y, z) in the space must be on the sphere with
radius r and center C(h,k,l).
6. Generally
very equation F(x, y, z) =0 with three variables
(x, y, z) represents a surface in space and vice versa.
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7. Example:
Discribe the surface which corresponds the equation
x 2+ y 2 + z2 +4x-6y-8z= 5.
8. Equation
Ax + By +Cz +D=0 (where A,B,C,D are constants)
represents a plane in three-dimensional space.
For instance, x=0 is yz-plane , z=5 is the plane parallel to and 5
units above xy-plane. How about y=10?
9. A special equation
z=f(x, y), if (x, y) is confined in a domain D,
means a piece of surface over or below the domain D.
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z
(x, y, f(x,y))
o
y
D
x
(x,y)
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10. The
curve is intersection of two surfaces F(x,y,z)=0 and
G(x,y,z)=0.
11. The intersection of two planes
A1x + B1y +C1z +D1=0 and
A2x + B2y +C2z +D2=0 is a straight line,or line for short.
12. Similarly curve in space also can be represented by parameter
equation: x=f(t), y=g(t), z=h(t).
13. Similarly curve in space also can be represented by parameter
equation: x=f(t), y=g(t), z=h(t).
And if the curve: x=f(t), y=g(t), z=h(t) (atb) is smooth, then
length of the curve is
L
b

[ f ' (t )]2  [ g ' (t )]2  [h' (t )]2 dt
a
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11.2 Quadratic surface
(1) Ellipsoid: The quadratic surface with equation
x2
y2
z2
 2  2 1
2
a
b
c
is called ellipsoid.
z
0 000 0
0.
0
0
0
0
0
00
0
x
y
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end
(2) Hyperboloid of one sheet:
x2
y2
z2
 2  2 1
2
a
b
c
z
y
HH H HHH
x
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end
(3) Hyperboloid of two sheets:
x2
y2
z2
 2  2  2 1
a
b
c
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end
(4) Cones:
x2
y2
z2
 2  2
2
a
b
c
z
y
x
HH H HHH
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end
(5) Elliptic paraboloid: The surface
x2
y2
z
 2 
2
c
a
b
Fig.
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end
(6) Hyperbolic paraboloid: The surface
x2
y2
z
 2 
2
c
a
b
Fig.
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end
(7) Elliptic cylinder: The surface
z
y
x2+y2=R2
x
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end
(8) Parabolic cylinder: The surface
z
y
o
y=x2
x
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end
11.3(chapter 12.1) functions of several variables
1. Definition Let DR2. A function f of two variables is
a rule that assigns to each ordered pair (x,y) in D a unique
real number denoted by f (x,y) . The set D is the domain of f
and its range is the set of value that f takes on, that is ,{f (x,y)
| (x,y)D}.
From the definition,we write z= f (x, y) to make explicit
the value taken by f at general point (x,y). The variables
x and y are independent variables and z is the dependent
variable. Here D is represented as a subset of the xyplane.Actually z= f (x, y) is a surface in space.
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2. Example Find the domain of the function f (x,y)=xln( y2-x)
and evaluate f (3,2).
Solution Since f (x, y)=xln( y2-x) is defined only
when y2-x >0,that is , y2>x ,the domain of f is
D={(x,y)| y2>x }.This is a set of points to the left of
parabola y2=x .
And f (3,2)=3ln(22-3)=0
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3. Definition If is a function f of two variables with
domain D, the graph of f is the set:
S={(x,y,z)R3 | z=f (x,y) , (x,y)D}.
Actually graph of a function of two
variables is a surface.
z
For instance, z= -2x4y+1 is a plane through
A(0,0,1), B(0.5,0,0),and
C(0,0.25,0).
A(0,0,1)
C(0,0.25,0)
B(0.5,0,0)
o
y
x
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Tell what are the surfaces of the following functions:
x2
y2
z 2  2
a
b
2
x2
y2
z 2  2
a
b
2
x
y
z   1 2  2
a
b
x2
y2
z   1 2  2
a
b
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4. Definition The level curve of a function f of
two variables is the curve with equation f( x,y) =k,
where k is a constant (in the range of f).
From the definition, we know that the level curve of
a function f of two variables is vertical projection to
xy-plane of intersection of surface z = f( x,y) and
plane z =k, where k is a constant (in the range of f).
Examplez 
x2
y2
1 2  2
a
b
is upper ellipsoid,
whose domain is ellipse
x2
y2
 2 1
2
a
b
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en
11.4 (chapter 12.2) Limit and continuity
1. Definition
Let f be a function of two variables
defined a domain which contains a disk with center (a,b).
Then we say that the limit of f (x, y) as (x, y) approaches
(a, b) is L and we write
lim
( x , y )( a ,b )
f ( x, y )  L
or f(x,y) L
or
as
lim f ( x, y )  L
x a
y b
(x,y) (a,b).
If for every number >0 there is a corresponding
number >0 such that |f (x,y) - L |<  whenever
0
( x  a ) 2  ( y  b) 2  
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In other word, f(x,y) approaches L as (x,y) approaches
(a,b) along any path.
So, if there are two paths C1 and C2 , L1 and L2 ,
L1 L2 , f(x,y) approaches L1 as (x,y) approaches (a,b)
along C1,and f(x,y) approaches L2 as (x,y) approaches
(a,b) along C1, then
lim
( x , y )( a ,b )
f ( x, y ) does not exists.
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Example Find
if it exists.
lim
xy 2
x2  y 2
, does it exist?.
( x , y ) ( 0 , 0 )
Example Find
Example
lim
x2  y 2
x2  y 2
( x , y ) ( 0 , 0 )
xy 2
If f ( x, y )  2
x  y4
lim
( x , y ) ( 0 , 0 )
xy 2
x2  y 4
, does the limit
exist?
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2. Definition Let f be a function of two variables
defined on a disk with center (a,b). Then f is called
continuous at (a,b), if
lim
( x , y )( a ,b )
f ( x, y )  f ( a , b )
Definition Let D be subset of R2 , and (a,b)R2 , if there
exists disk D with center at (a,b) and radius >0, that is,
D ={(x,y) | (x - a)2+ (y - b)2 < 2 }, such that D D, then
(a,b) is called an interior point of D. If for every >0, D
contains points in D and also points not in D, then (a,b) is
called the boundary point of D.
If (a,b) is a boundary point of D, then
lim
( x , y )( a ,b )
f ( x, y )  L
means that for every number >0, there is a
corresponding number >0 such that |f (x,y) - L |< 
whenever (x,y)D and 0  ( x  a)2  ( y  b)2   .
So if f (x,y) is a function on DR2, (a,b) in D is a
boundary point of D, and lim f ( x, y )  f (a, b)
( x , y )( a ,b )
then we also call f (x,y) is continuous at (a,b) . If f (x,y)
is continuous at all points in D,then we say f (x,y) is a
continuous on D.
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From the discussion above, it is easy to get that sums,
differences, products, and quotients(except the points at
which denominator is equal to 0) of continuous functions
on their domain are continuous. So Polynomial function,
which is a sum of terms of the form cxmyn where c is a
constant and m and n are nonnegative integers, is a
continuous on R2, and Rational function, which is the
quotient of two polynomials, is continuous on R2 but the
points at which denominator is equal to 0.
Example Evaluate
lim ( xy  x 2 y 3  3x  2 y)
( x , y )( 2, 3)
x2  y 2
Example Where is the function f(x,y) = 2
x  y2
continuous?
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