CalcNotes0207-page4

CalcNotes0207-page4 - 4, 5 ( ) . This would have been the...

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(Section 2.7: Precise Definitions of Limits) 2.7.4 How Does the Static Approach to Limits Meld with the Dynamic Approach? Why is lim x ± 4 7 ² 1 2 x ³ ´ µ · ¸ = 5 ? Because, regardless of how small we make the tolerance level ± and how tight we make the lottery for the players, there is a value for for which the corresponding “symmetric punctured interval” is a winning one; in other words, the corresponding shaded region traps the graph of f on the symmetric punctured interval. As ² 0 + , we can choose values for in such a way that the shaded region always traps the graph and zooms in, or collapses in, on the point
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Unformatted text preview: 4, 5 ( ) . This would have been the case even if that point alone had been deleted from the graph. If = 1 , we can choose = 2 . If = 0.5 , we can choose = 1 . (Figure 2.7.e) (Figure 2.7.f) More generally for this example, if is any positive real constant, we can choose = 2 . Why is that? Graphically, we can exploit the fact that the slope of the line y = 7 1 2 x is 1 2 . Remember: slope = rise run . Along the line, an x-run of 2 units corresponds to a y-drop of 1 unit. We will now demonstrate this rigorously....
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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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