15.2 Limits and Continuity
1.Definition (Limit of f (x, y) as (x, y) → (a, b) )
Let f (x, y) be a function whose domain is D. Suppose (a, b) is a
point in D such that a disk centered at (a, b) is contained in D.
We say that the limit f (x, y) as (x, y) approaches (a, b) is L and
write
lim
f (x, y) = L
(x,y)→(a,b)
if for any possible path to approach (a, b) in D, f (x, y) approaches
unique number L.
(e.g) Given a function f (x) =
(a)
lim
f (x, y)
(x,y)→(0,0)
(b)
1 if
0 if
x≥y
, evaluate the limits :
x<y
lim
(c)
f (x, y)
(x,y)→(2,−2)
lim
f (x, y)
(x,y)→(1,1)
(d)
lim
(x,y)→(2,5)
(Answers) (The figure shows the graph of f (x, y) in the first octant.)
(a)
(b)
(c)
(d)
DNE
1
DNE
0
2.Example Show that the limit does not exist :
(a)
(Answers)
y2
(x,y)→(0,0) x2 + xy
(b)
lim
xy 4
(x,y)→(0,0) x5 + y 5
lim
(a) Let (x, y) approach (0, 0) along the line y = mx in xy−plane where m can be any real number. Then
lim
(x, y) → (0, 0)
y = mx
x2
y2
(mx)2
m2 x2
m2
= lim 2
= lim
=
2
x→0
x→0
+ xy
x + x(mx)
(m + 1)x
m+1
That is, depending on the choice of path(the choice of m), the limit result varies. For example,
1
if (x, y) approaches (0, 0) along the line y = x, the limit value is . On the other hand, if
2
4
(x, y) approaches (0, 0) along the line y = 2x, the limit value is . Hence, the limit as
3
(x, y) → (0, 0) does not exist.
(b) Similarly as in (a), let (x, y) approach (0, 0) along the line y = mx. Then
lim
(x, y) → (0, 0)
y = mx
xy 4
x(mx)4
m 4 x5
m4
= lim 5
= lim
=
5
5
5
5
x→0
x→0
+y
x + (mx)
(m + 1)x
m5 + 1
x5
As before, the limit value will vary as m changes. Hence, the limit does not exist.
3.Example Show that the limit exists and evaluate it :
1
f (x, y)
(a)
(Answers)
x4 − y 4
(x,y)→(0,0) x2 + y 2
(b)
lim
(a) Notice that if (x, y) 6= (0, 0),
xy 4
(x,y)→(0,0) x2 + y 4
lim
(x2 + y 2 )(x2 − y 2 )
x4 − y 4
2
2
=
= x − y . Hence
x2 + y 2
x2 + y 2
lim
(x,y)→(0,0)
x4 − y 4
=
x2 + y 2
2
2
x −y =0
lim
(x,y)→(0,0)
(b) For all real values x and y in the domain of the function f (x, y) = xy 4 /(x2 + y 4 ),
−
xy 4
y4
−x
−
lim
x
≤
xy 4
x2 + y 4
≤
≤
xy 4
x2 + y 4
≤x
≤
xy 4
x2 + y 4
(x,y)→(0,0)
≤
lim
xy 4
y4
x
(x,y)→(0,0)
By the Squeeze Theorem, as in part(a), the limit exists and equals 0.
4.Definition (Continuity) A function f (x, y) is said to be continuous at a point (a, b) if
•
lim
(x,y)→(a,b)
f (x, y) exists,
• f (a, b) exists,
• and
lim
f (x, y) = f (a, b)
(x,y)→(a,b)
A function f (x, y) is said to be continuous if it is continuous at every point (x, y) in the domain.
5.Theorem
(a) Function f (x, y) = Cxm y n (m, n = 0, 1, 2, · · · and C is a constant) is continuous.
(b) If f (x, y) and g(x, y) are continuous, then the following functions are continuous as well :
f (x, y) − g(x, y),
f (x, y) + g(x, y),
Furthermore, if g(x, y) 6= 0 at (a, b), then
f (x, y)g(x, y)
f (x, y)
is continuous at (a, b).
g(x, y)
(c) If f (x, y) is continuous at (a, b) and h(t) is continuous at f (a, b), then h(f (x, y)) is continuous at (a, b).
6.Example Find all points (x, y) at which f (x, y) is continuous.
(a) f (x, y) = x2 y − xy 2 + 3
(b) f (x, y) =




x2 y 3
2x2 + y 2
if
(x, y) 6= (0, 0)



1
if
(x, y) = (0, 0)
(c) f (x, y) = g(h(x, y)) where h(x, y) = 2x + 3y − 6 and g(t) = t2 + ln t.
2
(Answers)
(a) All (x, y) in xy−plane.
(b) All (x, y) in xy−plane except (0, 0), because
lim
(x,y)→(0,0)
x2 y 3
= 0 6= 1 = f (0, 0)
2x2 + y 2
(c) f (x, y) = (2x+3y−6)2 +ln(2x+3y−6) is well defined only when 2x+3y−6 > 0.
2
2
That is y > − x + 2 and it is the region on xy−plane above the line y = − x + 2.
3
3
See the figure on the right.
((x, y) on which f (x, y) in (c) is continuous)
7.Example Use continuity to evaluate the limit :
(a)
(b)
(Answers)
lim
e−xy cos(x + y)
(x,y)→(1,−1)
x2 + sin2 y
(x,y)→(0,0) 2x2 + y 2 + 1
lim
(a) e−xy cos(x + y) is continuous at (1, −1). Hence
(b)
lim
e
−xy
cos(x + y) = e
−1·(−1)
(x,y)→(1,−1)
x2 + sin2 y
x2 + sin2 y
0+0
is continuous at (0, 0. Hence
lim
=
= 0.
(x,y)→(0,0) 2x2 + y 2 + 1
2x2 + y 2 + 1
0+0+1
3
cos(1 + (−1)) = e