Limits and continuity for (Sect. 14.2) The limit of ...
Limits and continuity for f : Rn R (Sect. 14.2)
The limit of functions f : Rn R. Example: Computing a limit by the definition. Properties of limits of functions. Examples: Computing limits of simple functions. Continuous functions f : Rn R. Computing limits of non-continuous functions:
Two-path test for the non-existence of limits. The sandwich test for the existence of limits.
The limit of functions of several variables.
Definition
The limit of the function f : D Rn R, with n N, at the point P^ Rn is the number L R, denoted as lim f (P) = L,
P P^
iff for every > 0 there exists > 0 such that 0 < |P - P^ | < |f (P) - L| < .
Remarks:
(a) In Cartesian coordinates P = (x1, ? ? ? , xn), P^ = (x^1, ? ? ? , x^n). Then, |P - P^ | is the distance between points in Rn, - |P - P^ | = |PP^ | = (x1 - x^1)2 + ? ? ? + (xn - x^n)2.
(b) |f (P) - L| is the absolute value of real numbers.
The limit of functions f : R2 R.
Idea of the limit definition: Consider f : R2 R. Then,
lim f (x, y ) = L
(x,y )(x0,y0)
means that the closer (x, y ) is to (x0, y0) then the closer the value of f (x, y ) is to L.
z
L f(x,y)
y
(x 0,y 0)
x
Limits and continuity for f : Rn R (Sect. 14.2)
The limit of functions f : Rn R. Example: Computing a limit by the definition.
Properties of limits of functions.
Examples: Computing limits of simple functions. Continuous functions f : Rn R. Computing limits of non-continuous functions:
Two-path test for the non-existence of limits. The sandwich test for the existence of limits.
Computing a limit by the definition
Example
2yx 2
Use
the
definition
of
limit
to
compute
lim
(x,y )(0,0)
x2
+
y2.
2yx 2 Solution: The function f (x, y ) = x2 + y 2 is not defined at (0, 0). First: Guess what the limit L is.
0 Along the line x = 0 the function is f (0, y ) = y 2 = 0. Therefore, if L exists, it must be L = 0. Given , find . Fix any number > 0. Given that , find a number > 0 such that
0 < (x - 0)2 + (y - 0)2 <
2yx 2 x2 + y2 - 0 < .
Computing a limit by the definition
Example
2yx 2
Use
the
definition
of
limit
to
compute
lim
(x,y )(0,0)
x2
+
y2.
Solution: Given any > 0, find a number > 0 such that
Recall: x2
x2 + y2 <
2yx 2 x2 + y2 < .
x2 + y2,
that
is,
x2 x2 + y2
1. Then
2yx 2
x2
x2 + y 2 = (2|y |) x2 + y 2
2|y | = 2 y 2
2 x2 + y2.
Choose = /2. If
x2 + y 2 < , then
2yx 2 x2 + y2
< 2 =
.
We conclude that L = 0.
Limits and continuity for f : Rn R (Sect. 14.2)
The limit of functions f : Rn R. Example: Computing a limit by the definition. Properties of limits of functions. Examples: Computing limits of simple functions. Continuous functions f : Rn R. Computing limits of non-continuous functions:
Two-path test for the non-existence of limits. The sandwich test for the existence of limits.
Properties of limits of functions
Theorem
If f , g : D Rn R, with n N, satisfying the conditions
lim f (P) = L and lim g (P) = M, then holds
P P^
P P^
(a) lim f (P) ? g (P) = L ? M;
P P^
(b) If k R, then lim kf (P) = kL;
P P^
(c) lim f (P) g (P) = LM;
P P^
f (P) L
(d) If M = 0, then lim
=.
PP^ g (P) M
(e) If k Z and s N, then lim f (P) r/s = Lr/s .
P P^
Remark: The Theorem above implies: If f = R/S is the quotient
of two polynomials with S(P^ ) = 0, then lim f (P) = f (P^ ).
P P^
Limits and continuity for f : Rn R (Sect. 14.2)
The limit of functions f : Rn R. Example: Computing a limit by the definition. Properties of limits of functions. Examples: Computing limits of simple functions. Continuous functions f : Rn R. Computing limits of non-continuous functions:
Two-path test for the non-existence of limits. The sandwich test for the existence of limits.
Limits of R/S at P^ where S(P^ ) = 0 are simple to find
Example
x2 + 2y - x
Compute lim
.
(x,y )(2,1)
x -y
Solution: The function above is a rational function in x and y and its denominator is defined and does not vanish at (2, 1). Therefore
x2 + 2y - x 22 + 2(1) - 2
lim
=
,
(x,y )(2,1)
x -y
2-1
that is,
x2 + 2y - x
lim
= 4.
(x,y )(2,1)
x -y
................
................
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