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CalcNotes0206-page9

CalcNotes0206-page9 - x> so none of the inequality...

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(Section 2.6: The Squeeze (Sandwich) Theorem) 2.6.9 PART C: MODIFICATIONS FOR “LONG-RUN” LIMITS Example 4 (Revisiting Example 5 from Section 2.3) Find lim x ±² sin x x , and prove it. Solution Show how sin x is bounded. ± 1 ² sin x ² 1 , for all real x ± We may assume x > 0 in our analysis, since we are considering a limit as x ±² . Divide all three parts by x 2 so that the middle part becomes the expression we want to take the limit of. We assume
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Unformatted text preview: x > , so none of the inequality symbols have to be reversed. ± 1 x ² sin x x ² 1 x , if x > ± Now, we apply the Squeeze (Sandwich) Theorem. lim x ±² ³ 1 x ´ µ ¶ · ¸ ¹ = , and lim x ±² 1 x = . Therefore, lim x ±² sin x x = . Shorthand: As x ± ² , ³ 1 x ± ± ´ sin x x Therefore, ± ± ´ 1 x ± ± x > ( ) (See the graph on the next page.)...
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