2 and 01 we get aka f 1 b lim h 0 h k lim h0

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Unformatted text preview: k −1 or f (a + k ) = b + h. (b) = a and f (a) = b, by (0.2) and (0.1), we get a+k−a (f −1 )￿ (b) = lim h→ 0 h k = lim . h→0 f (a + k ) − f (a) Notice that, as h approaches zero, k = f −1 (b + h) − a also approaches zero, since lim (f −1 (b + h) − a) = f −1 (lim (b + h)) − a = f −1 (b) − a = 0, h→ 0 h→ 0 where we have used the continuity of f −1 in the first equality above. Note that f −1 is continuous, since f is continuous. Therefore, we can rewrite (0.3) as k (f −1 )￿ (b) = lim k → 0 f ( a + k ) − f ( a) 1 f ( a + k ) − f ( a) =￿ since lim = f ￿ ( a) , k →0 f ( a) k 1 = ￿ −1 . f (f (b)) ￿ 1...
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This note was uploaded on 02/10/2014 for the course MATH 235 taught by Professor Lucas during the Fall '08 term at James Madison University.

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