Multivariate Calculus - Bruce E. Shapiro

70 LECTURE 10. LIMITS AND CONTINUITY
We can get a general definition of a limit (in any dimension) by merely omitting the
language specific to 2D.
Definition 10.3 Let f(x) : R m ↦→ R n , let P be a point in R m , and L a point in
R n . Then the limit of f(x) as x approaches P, which we write as
lim f(x) = L
x→P
exists if and only if for any ɛ > 0 there exists some δ > 0 such that whenever 0 <
‖f(x) − (P)‖ < δ (i.e., x is in a neighborhood of radius δ of P) then ‖f(x) − L‖ < ɛ
(i.e., f(x) is in some neighborhood of radius ɛ of L).
Example 10.1 Calculate
as you approach the origin
x 2 − y 2
lim
(x,y)→(0,0) x 2 + y 2
(a) along the x-axis;
(b) along the line y=3x;
(c) along the parabola y = 5x 2 .
Solution.
(a) along the x-axis we have y=0, and we can approach the origin by letting x → 0.
Therefore
x 2 − y 2
lim
(x,y)→(0,0),y=0 x 2 + y 2 = lim x 2 − 0 2
x→0 x 2 + 0 2 = lim x 2
x→0 x 2 = lim 1 = 1
x→0
(b) Along the line y = 3x we have
x 2 − y 2
lim
(x,y)→(0,0),y=3x x 2 + y 2 = lim x 2 − (3x) 2
x→0 x 2 + (3x) 2 = lim −8x 2
x→0 10x 2 = −4 5
(c) Along the parabola y = 5x 2 we have
x 2 − (5x 2 ) 2 x 2 − 25x 4
lim
(x,y)→(0,0),y=x 2 x 2 + (5x 2 ) 2 = lim
x→0 x 2 + 25x 4 = lim x 2 (1 − 25x 2 )
x→0 x 2 (1 + 25x 2 )
= lim
x→0
1 − 25x 2
1 + 25x 2 = 1
The first and third limits are the same, but the third limit is different. Therefore
the limit does not exist.
As the above example shows, to show that a limit does not exist we need to
calculate the limit along different approach paths and show that different numbers
result. Showing that limit does exist is considerably more difficult than showing
that it does not exist.
Revised December 6, 2006. Math 250, Fall 2006