Mth132SupEx2-4FS10 - Supplemental Exercises for Section 2.4...

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Supplemental Exercises for Section 2.4 The formula lim x 0 sin x x = 1 can be made more useful by observing that if lim x a f ( x ) = 0 and f ( x ) is never 0, then lim x a sin f ( x ) f ( x ) = 1 . For example 1. lim x 0 sin x 2 x 2 = 1 . 2. lim x 0 sin x 2 x = lim x 0 sin x 2 x 2 lim x 0 x 2 x = 1 · 0 = 0 . 3. lim x →- 1 sin( x 2 - x - 2) x + 1 = lim x →- 1 sin( x 2 - x - 2) ( x 2 - x - 2) lim x →- 1 ( x 2 - x - 2) x + 1 = 1 · lim x →- 1 ( x + 1)( x - 2) x + 1 = - 3 . 4. lim x 1 sin(1 - x ) x - 1 = lim x 1 sin(1 - x ) 1 - x 1 - x x - 1 = 1 · lim x 1 (1 - x )(1 + x ) (1 + x )(1 + x ) = . lim x 1 1 - x ( x - 1)(1 + x ) = - 1 2 . Find each of the following limits.
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This note was uploaded on 10/03/2011 for the course MTH 132 taught by Professor Kihyunhyun during the Spring '10 term at Michigan State University.

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