Thus we 0 0 2 2 2 0 0 prove

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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.

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