Super Squeeze

# Super Squeeze - Theorem 0.1 Suppose f g h(c d → R and a...

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Unformatted text preview: Theorem 0.1. Suppose f, g, h : (c, d) → R and a ∈ R satisfy the following three conditions: (1) f (x) ≤ g (x) ≤ h(x); (2) f (a) = h(a); (3) f and h are diﬀerentiable at a with f (a) = h (a). Then g (a) exists and equals f (a) = h (a). Proof. For suﬃciently small (i) > 0, we know that a + ∈ (c, d) since a ∈ (c, d). By f (a + ) ≤ g (a + ) ≤ h(a + ) for such . By (ii) f (a + ) − f (a) ≤ g (a + ) − g (a) ≤ h(a + ) − h(a). Dividing by we have f (a + ) − f (a) ≤ g (a + ) − g (a) →0 f (a+ )−f (a) →0 ≤ h(a + ) − h(a) . h(a+ )−h(a) By (iii) we know that f (a) = lim Q.E.D. exists and equals h (a) = lim By the squeeze theorem, g (a) = lim g (a+ )−g (a) →0 . exists and equals f (a) = h (a). 1 ...
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## This note was uploaded on 06/26/2010 for the course MATH 1A taught by Professor Wilkening during the Fall '08 term at Berkeley.

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