5 e consider the function 2 2 2 2 0 2 2 2 0 0 prove

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Unformatted text preview: ∈ R, then the limit of its restriction on any curve would also be . However, here we found at least two curves — straight line = 0 and parabola = 2 such that the respective restrictions have different limits. Therefore lim( )→(0 0) ( ) is not defined. :) 5 E Consider the function ( 2 2 )= 2 2 + 0 2 + + 2 2 =0 =0 Prove that is continuous with respect to for each fixed = , continuous with respect to for each fixed = , but discontinuous at the point (0 0). S For each fixed = , if = 0, then by definition, 2 ( )= 2 2 + It’s easy to see that ( ) is a rational function of with non-zero denominator and therefore we know it’s continuous with respect to from calculus of one variable. If = 0, then it’s easy to see from the definition that ( 0) = 0 so it’s also continuous. Similarly, for each fixed = , if = 0, then by definition, ( 2 2+ )= 2 It’s easy to see that ( ) is a rational function of with non-zero denominator and therefore we know it’s continuous with respect to from calculus of one variable. I...
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This document was uploaded on 02/10/2014.

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