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Unformatted text preview: →(0 0) However, the limit of this expression along the line
conclude that is not diﬀerentiable at (0 0).
E Let √
( is 0 ( +
1
√,
2 2 =0 which is not 0. Thus we
:) ) = (0 0) ( 2+ 2 )= = 2  ) = (0 0) Prove that
(a) ( ) is continuous on R2 (including (0 0)); (b) Partial derivatives and are deﬁned on R2 (including (0 0)); (c) Partial derivatives and are bounded; (d) However, is not diﬀerentiable at (0 0). 5 S (a) First note that (
) is an elementary function which is well deﬁned at
all the points besides the origin. Hence (
) is continuous at all the points except
possibly at (0 0). To see whether (
) is continuous at (0 0) or not, we need to
check that lim( )→(0 0) (
) = 0. We have already seen that a function has limit
zero if and only if its modulus has limit zero. Hence we can prove continuity at the
origin by showing that lim( )→(0 0)  (
) = 0. Now, for (
) = (0 0) we have
0≤ ( ) = 2 = 2 + )·(
(
) (
 · 
≤
(
) ) ) →0 =( ) → (0 0) as (
Hence, by the Squeeze Theorem, the limit is 0. Thus, (
(0 0).
(b) If ( ) is also continuous at ) = (0 0), we have
( 2 − )= 2 2 + ( At the point (0 0), we have ( 2 + 2 )3/2 ( 0) ≡ 0 and (0 2 − )= 2 2 + ) ≡ 0 and hence ( 2 + (0 0) = 2 )3 / 2 (0 0) = 0. (c) It’s enough to estimate
since (
)= (
) and hence the case for
is completely similar. First note that we have (0 0) = 0 is bounded. For other points, we
have,
2 −
2 + 2 ( 2 + Thus we conclude that 2 )3 / 2 ≤1+ ·
2 + 2 ·
2 + 2 2 + 2 ≤2 is bounded by 2. (d) For diﬀerentiability, we must have
( ) − (0 0) − lim
( )→(0 0) (0 0) −
2 + 2 (0 0)
=0 Substituting the values we found, we get the limit lim( )→(0 0) 2 +  2 , which is easily
seen to be undeﬁned.Hence the function is not diﬀerentiable at (0 0). 6...
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This document was uploaded on 02/10/2014.
 Spring '13

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