MoreSols - x-c we can write for all x ∈ A with x 6 = c g...

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Additional Solutions, in Preparation for Exam #2 Exercise 4.2.3 (a) Consulting Corollary 4.2.5, what we need is two sequences ( x n ) and ( y n ), both converging to 0, with x n 6 = 0 and y n 6 = 0, and for which lim | x n | x n and lim | y n | y n disagree. For these we can take x n = 1 n , y n = - 1 n . Then ± | x n | x n ² = (1) 1 while ± | y n | y n ² = ( - 1) → - 1. So, by the corollary, lim x 0 | x | x does not exist. Proof of Theorem 5.2.5 I had a request to go over this. It is in the book, but I did it a little differently in class. Here is the way I did it. Proof. We assume that f : A R , f ( A ) B , g : B R , f is differentiable at c A , and g is differentiable at f ( c ). Define a new function h : B R by h ( y ) = ( g ( y ) - g ( f ( c )) y - f ( c ) for y 6 = f ( c ) g 0 ( f ( c )) for y = f ( c ) . Since, by definition of g 0 ( f ( c )), lim y f ( c ) h ( y ) = h ( f ( c )), we see that h is continuous at f ( c ). Moreover, we have g ( f ( x )) - g ( f ( c )) = h ( f ( x )) · ( f ( x ) - f ( c )) for all x A. (Both sides are 0 if f ( x ) = f ( c ).) Dividing both sides of this by
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Unformatted text preview: x-c we can write, for all x ∈ A with x 6 = c , g ( f ( x ))-g ( f ( c )) x-c = h ( f ( x )) f ( x )-f ( c )) x-c . Now f ( x ) is continuous at c because it is differentiable there, g ( y ) is continuous at y = f ( c ) for the same reason, so by Theorem 4.3.9 h ( f ( x )) is continuous at x = c : lim x → c h ( f ( x )) = h ( f ( c )) = g ( f ( c )). We can now take the limit as x → c , using the Algebraic Limit Theorem (Corollary 4.2.4): lim x → c g ( f ( x ))-g ( f ( c )) x-c = lim x → c h ( f ( x )) lim x → c f ( x )-f ( c )) x-c = g ( f ( c )) f ( c ) . This completes the proof....
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This note was uploaded on 11/07/2009 for the course MATH 3224 at Virginia Tech.

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