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Unformatted text preview: 5), consider the function f : R2 → R deﬁned by
�2
xy
(x, y ) = (0, 0)
� 2
4,
f (x, y ) = x +y
(2.10)
0,
x = y = 0. Claim. The directional derivative Du f (0) exists for all u. Proof. Let u = (h, k ). Then lim
t→0 f (tu) − f (0)
f (tu) = lim
t→0
t
�t 3 2 �
t hk
1
= lim 2 2
4k4
t→0 t h +t
t
�
0,
h=0
=
2
k /h, h �= 0. (2.11) So the limit exists for every u.
However, the function is a nonzero constant on a parabola passing through the
4
origin: f (t2 , t) = 2tt4 = 1 , except at the origin where f (0, 0) = 0. The function f
2
is discontinuous at the origin despite the existence of all directional derivatives.
• Guess 3. This guess will turn out to be correct.
Remember than in onedimensional calculus we deﬁned
f (a + t) − f (a)
, t→0 t f � (a) = lim (2.12) for a function f : I → R and a point a ∈ I . Now consider the function
λ : R → R deﬁned by
λ(t) = f � (a)t. (2.13) 2 Then,
f (a + t) − f (a) − λ(t)
f (a + t) − f (a)
− f � (a)
= lim
t→0
t→0
t
t
= 0. lim (2.14) So, λ(t) ≈ f (a + t) − f (a) when t is small. Now we generalize to n dimensions. Deﬁnition 2.3. Given an open subset U of Rn...
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This note was uploaded on 10/02/2010 for the course MAT unknown taught by Professor Unknown during the Fall '04 term at MIT.
 Fall '04
 unknown
 Calculus

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