09.16 - δ ) | f ( x ) – 3| < ε ” game? ..?. . Adam...

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Section 1.3: The notion of limit (continued) [Discuss Example 9 on page 32 two ways: in terms of the picture, and in terms of strategies for Adam and Eve.] The limit of f ( x ) as x approaches a may be undefined, e.g., lim x 0 1/ x . The existence of the limit, and the value of the limit, do not depend on the value of f ( x ) at x = a (the value f ( a ) may even be undefined!) Example: Let f ( x ) = 2 for x 1, with f (1) = 3. [Draw a picture.] What is lim x 1 f ( x )? ..?. . lim x 1 f ( x ) = 2, even though f (1) = 3. To see why this is true, we go back to the definition of limit. Who wins the “(for all ε > 0) (there exists δ > 0 such that) (for all x with 0 < | x –1| < ) | f ( x ) – 2| < ” game? ..?. . Eve wins, regardless of what > 0 she picks! Let’s replace L = 2 by L = 3 to see what happens. Who wins the “(for all > 0) (there exists > 0 such that)
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Unformatted text preview: δ ) | f ( x ) – 3| < ε ” game? ..?. . Adam wins, by taking = 1 (or anything smaller). Moral: lim x → a f ( x ) need not equal f ( a ). Another example: Let f ( x ) = x +1 for all x ≠ 1, with f (1) = 3. [Draw a picture.] What is lim x → 1 f ( x )? ..?. . lim x → 1 f ( x ) = 2, even though f (1) = 3. Who wins the “(for all > 0) (there exists > 0 such that) (for all x with 0 < | x –1| < ) | f ( x ) – 2| < ” game? ..?. . Eve wins. What’s Eve’s strategy for winning the game? ..?. . = . Check: 0 < | x –1| < implies | f ( x ) – 2| = | x +1 – 2| = | x –1| < = . This verifies the claim “(for all > 0) (there exists > 0 such that) (for all x with 0 < | x –1| < ) | f ( x ) – 2| < ”....
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This note was uploaded on 02/13/2012 for the course MATH 141 taught by Professor Staff during the Fall '11 term at UMass Lowell.

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09.16 - δ ) | f ( x ) – 3| < ε ” game? ..?. . Adam...

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