CalcNotes0206-page4

CalcNotes0206-page4 - (Section 2.6: The Squeeze (Sandwich)...

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(Section 2.6: The Squeeze (Sandwich) Theorem) 2.6.4 If, say, a ± 4 , then ± 4 ² a ² 4 . Similarly, ± x 3 ² x 3 sin 1 x 3 ³ ´ µ · ¸ ² x 3 , if x ± 0 ± Now, we apply the Squeeze (Sandwich) Theorem. lim x ± 0 ² x 3 () = 0 , and lim x ± 0 x 3 = 0 . Therefore, lim x ± 0 x 3 sin 1 x 3 ² ³ ´ µ · = 0 . Shorthand: As x ± 0, ² x 3 ± 0 ±²³ ³ x 3 sin 1 x 3 ´ µ · ¸ ¹ Therefore, ± 0 ±² ´ ´³ ´´ ³ x 3 ± 0 µ x º 0 Solution 2 (Analyze Right-Hand and Left-Hand Limits Separately) Analyze: lim x ± 0 + x 3 sin 1 x 3 ² ³ ´ µ · . We may assume
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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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