Extra problems on partial derivatives e for the

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Unformatted text preview: f = 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.

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