CalcNotes0206-page5

CalcNotes0206-page5 - (Section 2.6: The Squeeze (Sandwich)...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
(Section 2.6: The Squeeze (Sandwich) Theorem) 2.6.5 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 ± ³ x 3 sin 1 x 3 ´ µ · ¸ ¹ Therefore, ± 0 ²³ ´ ´µ ´´ ³ x 3 ± 0 ± x > 0 Analyze: lim x ± 0 ² x 3 sin 1 x 3 ³ ´ µ · ¸ . We may assume x < 0 in our analysis, since we are considering a limit as x ± 0 ² . Show how sin 1 x 3 ± ² ³ ´ µ is bounded. ± 1 ² sin 1 x 3 ³ ´ µ · ¸ ² 1 , if x < 0 ± We will now multiply all three parts by
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online