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[CHAP. 6
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LIMITS Let f x; y be de ned in a deleted  neighborhood of x0 ; y0 [i.e.; f x; y may be unde ned at x0 ; y0 ]. We say that l is the limit of f x; y as x approaches x0 and y approaches y0 [or x; y lim lim f x; y l] if for any positive number  we approaches x0 ; y0 ] and write x!x f x; y l [or can nd some positive number  [depending on  and x0 ; y0 , in general] such that j f x; y lj <  whenever 0 < jx x0 j <  and 0 < j y y0 j < . If desired we can use the deleted circular neighborhood open ball 0 < x x0 2 y y0 2 < 2 instead of the deleted rectangular neighborhood.
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3xy if x; y 6 1; 2 . As x ! 1 and y ! 2 [or x; y ! 1; 2 ], f x; y gets closer to 0 if x; y 1; 2 3 1 2 6 and we suspect that lim f x; y 6. To prove this we must show that the above de nition of limit with EXAMPLE. Let f x; y l 6 is satis ed. Such a proof can be supplied by a method similar to that of Problem 6.4. Note that lim f x; y 6 f 1; 2 since f 1; 2 0. The limit would in fact be 6 even if f x; y were not de ned at 1; 2 . Thus the existence of the limit of f x; y as x; y ! x0 ; y0 is in no way dependent on the existence of a value of f x; y at x0 ; y0 .
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x!1 y!2 x!1 y!2
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approach of x; y to x0 ; y0 . It follows that if two di erent approaches give di erent values, the limit cannot exist (see Problem 6.7). This implies, as in the case of functions of one variable, that if a limit exists it is unique. The concept of one-sided limits for functions of one variable is easily extended to functions of more than one variable.
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EXAMPLE 1. lim tan 1 y=x =2, lim tan 1 y=x =2.
x!0 y!1
x;y ! x0 ;y0
f x; y to exist, it must have the same value regardless of the
x!0 y!1
EXAMPLE 2. lim tan 1 y=x does not exist, as is clear from the fact that the two di erent approaches of Example
x!0 y!1
1 give di erent results.
In general the theorems on limits, concepts of in nity, etc., for functions of one variable (see Page 21) apply as well, with appropriate modi cations, to functions of two or more variables.
CHAP. 6]
PARTIAL DERIVATIVES
ITERATED LIMITS The iterated limits lim
y!y0 x!x0 x!x0
y!y0
' & ' lim f x; y and lim lim f x; y , [also denoted by lim lim f x; y and
y!y0 x!x0 x!x0 y!y0
0 y!y0
lim lim f x; y respectively] are not necessarily equal. Although they must be equal if x!x f x; y is to lim
exist, their equality does not guarantee the existence of this last limit.
EXAMPLE. If f x; y
    x y x y x y , then lim lim lim 1 1 and lim lim lim 1 1. Thus x!0 y!0 x y x!0 y!0 x!0 x y y!0 x y the iterated limits are not equal and so lim f x; y cannot exist.
x!0 y!0
CONTINUITY Let f x; y be de ned in a  neighborhood of x0 ; y0 [i.e.; f x; y must be de ned at x0 ; y0 as well as near it]. We say that f x; y is continuous at x0 ; y0 if for any positive number  we can nd some positive number  [depending on  and x0 ; y0 in general] such that j f x; y f x0 ; y0 j <  whenever jx x0 j <  and jy y0 j < , or alternatively x x0 2 y y0 2 < 2 . Note that three conditions must be satis ed in order that f x; y be continuous at x0 ; y0 . 1.
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