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= 0, then it’s easy to see from the deﬁnition that (0 ) = 0 so it’s also continuous.
To show (
) is discontinuous at (0 0), we consider
lim (
0 This readily implies that ( 4 2
0 (1 + ) = lim 2
2) 2 2
1+ = 2 = 0 = (0 0) if =0 :) ) is discontinuous at (0 0). Extra problems on partial derivatives E
For the function (
) = 2 − + , prove that its limit at inﬁnity along any
straight line passing through the origin is 0. Also prove that, however, the limit along
the straight line = 1 as → +∞ is inﬁnite.
2 6 S Any straight line passing through the origin can be written as =
for some
∈ R except for the line = 0. On this line, we have (0 ) = 0 so that lim ∞ (0 ) = 0.
In all the other cases, we have
lim (
∞ Note that we have lim
twice to see that +∞ ∞ 2− 2 2+ = lim ∞ 2 ∞ 2− +∞ = lim 2 lim 2− ) = lim = +∞ so we can apply L’Hospital’s rule 2
= lim 2− (2 − ) ∞ = lim ((2 − ∞ which gives the desired result.
On the other hand, the limit along the straight line
is given by:
−12 + ) = lim 12 lim (1
∞ E 2− ∞ )2 2
+ 2) = 1 as
−1 = lim ∞ 2− =0 → +∞ of our function = +∞ :) Find the iterated limits
lim lim →∞ →0+ and lim lim →0+ →∞ 1+ +∞, we may assume S
For the ﬁrst limit, as the outer limit is
beginning so that we have for any ﬁxed > 1,
0 lim →0+ =
1+ 1+ 1+ 0 = > 1 at the 1
2 It follows that
1
1
=
→∞ 2
2 lim lim →∞ →0+ = lim
1+ 0+ , we may assume For the second limit, as the outer...
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This document was uploaded on 02/10/2014.
 Spring '13

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