Unformatted text preview: ) =
and (xδ − 0)2 + (yδ − 0)2 = δ2
2 = δ
√
2 <δ
δδ
f (xδ , yδ ) − f (0, 0) = f 2 , 2 − 0 = 1. This proves that f is not continuous in the origin. On the other hand, we have that f (0, y ) = x2 = 0 if y = 0, and f (0, y ) = 0
otherwise. So, f (0, y ) = 0 for every y ∈ R. Hence, the function y → f (0, y )
is continuous in y = 0. Analogously f (x, 0) = x2 , and, again, x → f (x, 0) is
continuous in x = 0.
Exercise 0.3. Let f : R2 → R be deﬁned as:
xy
if (x, y ) = (0, 0)
2
2
f (x, y ) := x +y
.
0
otherwise
Is f continuous in the origin? Why? 3 Directional derivatives and Diﬀerentiability
Let E ⊂ RN , let f : E → R and let x0 ∈ E . Given direction v ∈ RN , let L be
the line through x0 in the direction v, that is,
L := {x ∈ RN : x = x 0 + tv , t ∈ R }
1 and assume that x0 is an accumulation point of the set E ! L. The directional
derivative of f at x0 in the direction v is deﬁned as
!f
f (x0 + tv) " f (x0 )
(x0 ) := lim
,
t!0
!v
t
provided the limit exists in R. In the special case in which v = ei , the directional
!f
derivative ! ei (x0 ), if it exists, is called the p...
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 Spring '14
 Geometry

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