Download The fundamental philosophy of calculus is to a) approximate, b

The fundamental philosophy of calculus is to a) approximate, b) refine the approximation, and c) apply a limit process. In order to deal with c), we’ll need the
concept of a limit, specifically, for functions of several variables.
Definition 1. The limit as (x, y) approaches (a, b) of f(x, y) equals L, denoted
lim(x,y)→(a,b) f(x, y) = L, means that for all > 0, there exists δ > 0 such that if
p
0 < (x − a)2 + (y − b)2 < δ, then |f(x, y) − L| < . 1
In laymen’s terms, this means that f(x, y) can be made arbitrarily close to L, say
within > 0, provided that (x, y) is sufficiently close to (a, b), say within δ > 0.
Using the definition of the limit to prove that the value of a limit is L amounts to
being explicit about the relationship between and δ, i.e., how does δ depend on
.
2
y
x +y2
Example 1. Prove lim(x,y)→(0,0) √2x2
= 0.
Proof. We want to show for all > 0, there exists δ > 0 such that if 0 <
p
x2 + y2 <
δ, then |f(x, y)| < . So, we begin by fixing > 0. How does δ depend on ? We
begin by trying to find a “nice” upper bound for |f(x, y)|. Note
2x2 y 2x2 y p
p
· x2 + y2 6 2|y| x2 + y2 ,
p
= 2
2
x2 + y2 x +y
p
since x2 6 x2 + y2 . Also |y| 6 x2 + y2 . So
2x2 y p
2
p
(∗)
x2 + y2 .
6 2|y| x2 + y2 6 2
p
x2 + y2 p
p
Inspired from the above inequality, we set δ = /2 and assume 0 < x2 + y2 < δ.
2
p
2
p
x2 + y2 < δ2 = /2. That is, 2
x2 + y2 < . But we already know
Then
p
2
from (∗) that |f(x, y)| <
x2 + y2 . Putting everything together we get
2x2 y p
2
x2 + y2 < ,
p
62
x2 + y2 as desired.
2
y
x +y2
Example 2. Prove lim(x,y)→(0,0) √2x2
= 0 using the squeeze theorem, and the
theorem about which class of functions are continuous.
1The limit as (x, y) → (a, b) of f(x, y) doesn’t exist means for all L, there exists >
0 such that for all δ > 0 there exists (x, y) satisfying 0 <
|f(x, y) − L| > .
1
p
(x − a)2 + (y − b)2 < δ and
2
Proof. We already know from example 1, that
2x2 y p
2
0 6 p
x2 + y2 = P(x, y).
62
x2 + y2 The function P(x, y) is the composition of continuous functions, namely, we have
p
2
2
x2 + y2 = 2F(H(G(x, y))),
√
where F(t) = t2 (a power function), H(t) = t (a root function), and G(x, y) =
x2 + y2 (a polynomial function). We know that G is continuous at (0, 0), and H
and F are continuous at 0, and we also know that compositions of such functions
are continuous. It follows that P(x, y) is continuous at (0, 0). Since P(0, 0) = 0, the
2
squeeze theorem implies that lim(x,y)→(0,0) √2x2 y 2 = 0.
x +y
Problem 1. Show that lim(x,y)→(0,0)
x3 y3
x4 +y4
= 0. Hint: think of
x3 y3
x4 +y4
as a func-
tion of x, say f(x), where y is constant, then try to find an upper bound for f(x)
(depending on y, of course) on (−∞, ∞) using methods from single variable calculus.
In order to show limits don’t exist, we can use the following
Propisition 1. If there exists paths C1 and C2 such that
lim
f(x, y)
lim
f(x, y) 6=
(x,y)→(a,b)
C1
(x,y)→(a,b)
,
C2
then lim(x,y)→(a,b) f(x, y) does not exist.
This is the 2-dimensional analog of a familiar idea from single variable calculus,
namely limx→a− f(x) 6= limx→a+ f(x) implies limx→a f(x) does not exist.
Example 3. Show that lim(x,y)→(0,0)
x2 y3
x4 +y6
does not exist.
Solution. Let C1 be the x − axis. Then C1 is parametrized by x(t) = t and
y(t) = 0, moreover, t = 0 corresponds to (0, 0). So
x2 y3 t2 · 03
lim
=
lim
= 0.
t→0 t4 + 06
(x,y)→(a,b) x4 + y6 C
1
Now, let C2 be the curve y = x2/3 . Then C2 is parametrized by x(t) = t and
y(t) = t2/3 , and again, t = 0 corresponds to (0, 0). So
x2 y3 t2 · (t2/3 )3
t4
1
= lim 4 = .
lim
= lim 4
4
6
2/3
6
t→0 2t
t→0 t + (t
2
(x,y)→(a,b) x + y C
)
2
Since 1/2 6= 0, it must be that the limit does not exist.
3
(Challenge) Problem 2. Let
f(x, y) =
y −1/(x2 +y2 )
xe
x 6= 0
0
else.
Show that
(a)
lim(x,y)→(0,0) f(x, y)C = 0 where C is any straight-line path through the origin.
(b) the limit as (x, y) → (0, 0) of f(x, y) does not exist.