20e-lecture3 - Lecture 3 Friday April 3rd [email protected]

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Lecture 3 : Friday April 3rd [email protected] Key concepts : Derivative, Differentiability Know how to find derivative from definition and using usual methods 3.1 Further examples of non-existent limits Example 1. Prove that the limit as ( x, y ) 0 of f ( x, y ) = ( x + y + x 2 ) / ( x - y ) does not exist. Solution. Well let’s put y = mx to get f ( x, mx ) = (1 + m ) / (1 - m ) + x/ (1 - m ) for m 6 = 1. As x 0, this is (1 + m ) / (1 - m ) which depends on m . For example if m = 0 we get 1 whereas if m = - 1 we get 0. Therefore the original limit does not exist. The next example is important because it shows that trying lines y = mx to show that a limit as ( x, y ) (0 , 0) does not exist might fail. The point is, even if all limits lim x 0 f ( x, mx ) are equal (regardless of the value of m ) this does not mean lim ( x,y ) (0 , 0) f ( x, y ) exists. Instead, we try a different curve through (0 , 0), in the next example, the curve we try is y = m x for different values of m . Example 2. Prove that the limit of xy 2 / ( x 2 + y 4 ) as ( x, y ) 0 does not exist. Solution. If we put y = mx we get mx 3 / ( x 2 + m 4 x 4 ) = mx/ (1 + m 4 x 2 ). Now this limit as x 0 exists, so we can’t conclude that the original limit does not exist. Therefore we select a different path, say y = m x : in this case we get the function m 2 x 2 / ( x 2
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