=========================preview======================
(MATH100)worksheet-ch3-sol.pdf
Back to MATH100 Login to download
======================================================
MATH 100 (L1) Introduction to Multivariable Calculus Spring 2006-07
. Class Worksheet Chapter 3 Partial Derivatives
1. (Limits) Evaluate the limits (if they exist).
2x.y 2 4 cos xy

x 4 . 16y
(a) lim e . (g) lim .
(d) lim .
(x,y)(1,.3) (x,y)(0,0) x + y
(x,y)(0,0) x2 +4y2
xy
(b) lim xy 2 sin xy.2x 2 y
(e) lim .
(x,y)(1/2,)2 (h) lim .
(x,y)(0,0) x2 + y
(x,y)(0,0) x4 + y2 x 3 + y 3 x . y
(c) lim . (f) lim . (i) lim x ln(|x| + |y|).
(x,y)(0,0) x + y (x,y)(0,0) x2 + y2(x,y)(0,0)
Solution
(a) By direct observation,
22
2x.y 2(1).(.3)2.9 .7
lim e = e= e = e.
(x,y)(1,.3)
The limit exists.
(b) f(x, y)= xy 2 sin xy is continuous (everywhere), it follows that
1

lim f (x, y)= f( ,),
(x,y)(1/2,) 2
or

2 2
lim xy sin xy = ( 1)()2 sin( )= .
(x,y)(1/2,) 2 22
The limit exists.

(c) Note that f(x, y)=(x 3 + y 3)/(x + y) is de.ned at all points of the xy-plane except the origin (0, 0). We can still ask whether lim f(x, y) exists or not. The reason behind is whether
(x,y)(0,0)
f(0, 0) is de.ned 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) .= (0, 0), we can do cancellation,
33 2
x + y (x + y)(x 2 . xy + y )
lim = lim
(x,y)(0,0) x + y (x,y)(0,0) x + y = lim (x 2 . xy + y 2)
(x,y)(0,0)
= (0)2 . (0)(0) + (0)2 =0.
The limit exists.

(d) We can do cancellation,
4 222
x 4 . 16y (x 2 . 4y )(x +4y )
lim = lim
22
(x,y)(0,0) x2 +4y(x,y)(0,0) x2 +4y= lim (x 2 . 4y 2)
(x,y)(0,0)
= (0)2 . 4(0)2 =0.
The limit exists.

(e) Let f(x, y)= xy/(x 2 + y 2). Ifwelet (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 f (x, y) must be zero
(x,y)(0,0)
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 di.erent 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,
xy
lim does not exist.
(x,y)(0,0) x2 + y2
(f) The limit does not exist because we can .nd 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),
x . yx . 01

lim = lim =lim
2
(x,y)(0,0) x2 + y(x,0)(0,0) x2 +0 x0 x
which does not exist because both one-sided (left and right) limits do not exist:
11

lim = . and lim = .
x0. x x0+ x
(g) The limit does not exist because we can .nd a path along which the limit indeed does not exist.
Let (x, y) (0, 0) along the x-axis (y = 0),
cos xy cos0 1

lim =lim =lim
(x,y)(0,0) x + y x0 x +0 x0 x
which does not exist by the same reasoning in the previous