CalcNotes0206-page6

CalcNotes0206-page6 - (Section 2.6 The Squeeze(Sandwich...

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Unformatted text preview: (Section 2.6: The Squeeze (Sandwich) Theorem) 2.6.6 Now, we apply the Squeeze (Sandwich) Theorem. x ( x ) = 0. lim x 3 = 0 , and lim 0 x 3 0 1 Therefore, lim x 3 sin x 3 0 = 0. x Shorthand: x3 As x 0, x 3 sin 1 3 0 x3 x ( x < 0) 0 Therefore, 0 Now, we apply the Squeeze (Sandwich) Theorem. x lim+ 0 ( x ) = 0 , and 3 x lim+ x 3 = 0 . 0 1 Therefore, lim+ x 3 sin x 3 0 = 0. x Shorthand: x3 0+ , As x x 3 sin 0 1 3 x3 x ( x > 0) 0 Therefore, 0 Now, lim+ x 3 sin x 0 lim x 3 sin x 0 1 3 x 1 3 x = 0. = 0 , and lim+ x 3 sin x 0 1 3 x = 0 , and so ...
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This note was uploaded on 12/29/2011 for the course MATH 150 taught by Professor Sturst during the Fall '10 term at SUNY Stony Brook.

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