(MATH100)worksheet-ch3-sol - (MATH100)worksheet-ch3-sol.pdf...

MATH 100 (L1) Introduction to Multivariable Calculus Spring 2006-07
¥ Class Worksheet Chapter 3 Partial Derivatives
1. (Limits) Evaluate the limits (if they exist).
(a) lim
( x,y ) → (1 , - 3)
e
2 x - y
2
.
(b) lim
( x,y ) → (1 / 2 ,π )
xy
2
sin xy .
(c) lim
( x,y ) → (0 , 0)
x
3
+ y
3
x + y
.
(d) lim
( x,y ) → (0 , 0)
x
4
- 16 y
4
x
2
+ 4 y
2
.
(e) lim
( x,y ) → (0 , 0)
xy
x
2
+ y
2
.
(f) lim
( x,y ) → (0 , 0)
x - y
x
2
+ y
2
.
(g) lim
( x,y ) → (0 , 0)
cos xy
x + y
.
(h) lim
( x,y ) → (0 , 0)
2 x
2
y
x 4 + y 2
.
(i) lim
( x,y ) → (0 , 0)
x ln( | x | + | y | ).
Solution
(a) By direct observation,
lim
( x,y ) → (1 , - 3)
e
2 x - y
2
= e
2(1) - ( - 3)
2
= e
2 - 9
= e
- 7
.
The limit exists.
(b) f ( x, y ) = xy
2
sin xy is continuous (everywhere), it follows that
lim
( x,y ) → (1 / 2 ,π )
f ( x, y ) = f (
1
2
, π ) ,
or
lim
( x,y ) → (1 / 2 ,π )
xy
2
sin xy = (
1
2
)( π )
2
sin(
π
2
) =
π
2
2
.
The limit exists.
(c) Note that f ( x, y ) = ( x
3
+ y
3
) / ( x + y ) is defined at all points of the xy -plane except the origin
(0 , 0). We can still ask whether lim
( x,y ) → (0 , 0)
f ( x, y ) exists or not. The reason behind is whether
f (0 , 0) is defined or not has nothing to do with the existence of the limit. By existence of limits
we take care of the function values f ( x, y ) when the points ( x, y ) are close to (0 , 0) but not
exactly equal to (0 , 0). By this problem, since ( x, y ) 6 = (0 , 0), we can do cancellation,
lim
( x,y ) → (0 , 0)
x
3
+ y
3
x + y
= lim
( x,y ) → (0 , 0)
( x + y )( x
2
- xy + y
2
)
x + y
= lim
( x,y ) → (0 , 0)
( x
2
- xy + y
2
)
= (0)
2
- (0)(0) + (0)
2
= 0 .
The limit exists.
(d) We can do cancellation,
lim
( x,y ) → (0 , 0)
x
4
- 16 y
4
x
2
+ 4 y
2
= lim
( x,y ) → (0 , 0)
( x
2
- 4 y
2
)( x
2
+ 4 y
2
)
x
2
+ 4 y
2
= lim
( x,y ) → (0 , 0)
( x
2
- 4 y
2
)
= (0)
2
- 4(0)
2
= 0 .
The limit exists.
(MATH100)worksheet-ch3-sol.pdf downloaded by twuac from at 2014-02-26 20:42:36. Academic use within HKUST only.

(e) Let f ( x, y ) = xy/ ( x
2
+ y
2
). If we let ( x, y ) approach (0 , 0) along the x -axis ( y = 0), then
f ( x, y ) = f ( x, 0) → 0 (because f ( x, 0) = 0 identically). Thus, lim
( x,y ) → (0 , 0)
f ( x, y ) must be zero
if it exists at all. Similarly, at all points of the y -axis we have f ( x, y ) = f (0 , y ) = 0. However,
at points of the line y = x , f has a different constant value, f ( x, x ) = 1 / 2. Since the limit of
f ( x, y ) is 1 / 2 as ( x, y ) approaches (0 , 0) along this line, it follows that f ( x, y ) cannot have a
unique limit at the origin. That is,
lim
( x,y ) → (0 , 0)
xy
x
2
+ y
2
does not exist .
(f) The limit does not exist because we can find a path (yes, one path is good enough!) along which
the limit indeed does not exist. Let ( x, y ) → (0 , 0) along the x -axis ( y = 0),
lim
( x,y ) → (0 , 0)
x - y
x 2 + y 2
= lim
( x, 0) → (0 , 0)
x - 0
x 2 + 0
= lim
x → 0
1
x
which does not exist because both one-sided (left and right) limits do not exist:
lim
x → 0
-
1
x
= -∞ and lim
x → 0
+
1
x
= ∞ .
(g) The limit does not exist because we can find a path along which the limit indeed does not exist.
Let ( x, y ) → (0 , 0) along the x -axis ( y = 0),
lim
( x,y ) → (0 , 0)
cos xy
x + y
= lim
x → 0
cos 0
x + 0
= lim
x → 0
1
x
which does not exist by the same reasoning in the previous part.
(h) As in part (e), f ( x, y ) = 2 x
2
y/ ( x
4
+ y
2
) vanishes identically on the coordinate axes, so
lim
( x,y ) → (0 , 0)
f ( x, y ) must be 0 if it exists at all. If we examine f ( x, y ) at points along the
oblique straight lines y = kx , we obtain
f ( x, kx ) =
2 kx
3
x 4 + k 2 x 2
=
2 kx
x 2 + k 2
→ 0 , as x → 0 ( k 6 = 0) .