CalcNotes0206-page3

CalcNotes0206-page3 - 1 x 3 1 , if x , but we take for...

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(Section 2.6: The Squeeze (Sandwich) Theorem) 2.6.3 Example 2 Find lim x ± 0 x 3 sin 1 x 3 ² ³ ´ µ · , and prove it. Solution 1 (Using Absolute Value) Show how sin 1 x 3 ± ² ³ ´ µ is bounded. ± 1 ² sin 1 x 3 ³ ´ µ · ¸ ² 1 , if x ± 0 ± The problem with multiplying all three parts by x 3 is that, when x < 0 , x 3 < 0 . The inequality symbols would have to be reversed for negative values of x . One way to avoid this problem is to use absolute value. sin 1 x 3 ± ² ³ ´ µ · 1 , if x ± 0 ± Note : More precisely, we could write 0 ± sin
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Unformatted text preview: 1 x 3 1 , if x , but we take for granted that the absolute value of any real quantity is nonnegative. We will now multiply both sides of the inequality by x 3 . We know x 3 &gt; , if x . x 3 sin 1 x 3 x 3 , if x The product of absolute values equals the absolute value of the product. x 3 sin 1 x 3 x 3 , if x...
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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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