Lesson 10.pdf - 10.Limits of Functions Review for functions 1 A function f from a subset D ⊂ R into R is a rule which associates with each x ∈ D one

# Lesson 10.pdf - 10.Limits of Functions Review for functions...

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10.Limits of Functions July 19, 2019 Review for functions 1. A function f from a subset D R into R is a rule which associates with each x D one and only one y R . 2. Notation: f : D R . 3. D is called the domain of definition of f . 1 Limit of function After defining limit for a sequence, we are in a position to introduce the notion of limit for a function now. Definition Let f : D R where D is a subset of R and let c be an accumulation point of D . A number L is the limit of f at c if for any > 0, there exists δ > 0 such that | f ( x ) - L | < , x D, and 0 < | x - c | < δ. In this case, we denote lim x c f ( x ) = L , or f ( x ) L as x c . [Examples and remarks] 1. Geometrically lim x c f ( x ) = L means: “When x goes to c , f ( x ) goes to L .” 2. Recall x N ( c, δ ) if and only if c - δ < x < c + δ . Then the definition of lim x c f ( x ) = c is equivalent to: for any N ( L, ), there exists a corresponding deleted neighborhood N * ( c, δ ) such that f ( N * ( c, δ )) N ( L, ) . 1 For more information, see the section on functions in this textbook. 88

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3. Notice the part “0 < ” of the 0 < | x - c | < δ . This means that we do not require that the function f ( x ) is well defined at c . This is convenient. For example, we shall prove lim x 0 sin x x = 1 . Here the function f ( x ) = sin x x is not well defined at x = 0. 4. Note that we do not require D is an interval. Here D can be a sequence ( s n ) and f is a function defined on this sequence. Even D is an interval [ a, b ], the point c could be an end point a or b so that the above definition includes the one-side-limits (see the “one side limit” in this section). 5. Prove lim x 3 (5 x - 3) = 12. Proof: To show: for any > 0, there exists δ > 0 such that | (5 x - 3) - 12 | < , 0 < | x - 3 | < δ. (1) Consider | (5 x - 3) - 12 | = 5 | x - 3 | < . Then we take δ = 5 so that (1) holds. 6. Prove lim x 2 2 x 2 +4 x - 16 x - 2 = 12. Proof: To show: for any > 0, there exists δ > 0 such that 2 x 2 + 4 x - 16 x - 2 - 12 < , 0 < | x - 2 | < δ. (2) Consider 2( x 2 + 2 x - 8) x - 2 - 12 = 2( x - 2)( x + 4) x - 2 - 12 = | 2( x + 4) - 12 | = 2 | x - 2 | < . Then we take δ = 2 so that (2) holds. 7. Prove lim x 5 ( x 2 - 3 x + 1) = 11. Proof: To show: for any > 0, there exists δ > 0 such that | ( x 2 - 3 x + 1) - 11 | < , 0 < | x - 5 | < δ. (3) 89
Consider | ( x 2 - 3 x + 1) - 11 | = | x 2 - 3 x - 10 | = | ( x + 2)( x - 5) | = | x + 2 || x - 5 | < . (4) We let δ 1 = 1. Then 0 < | x - 5 | < δ 1 = 1 implies | x + 2 | = | ( x - 5) + 7 | ≤ | x - 5 | + 7 < δ 1 + 7 = 8. Then From (4), it is sufficient to consider 2 | ( x 2 - 3 x + 1) - 11 | = | x + 2 || x - 5 | < 8 | x - 5 | < . Then we take δ 2 := 8 . Finally we take δ := min { δ 1 , δ 2 ) so that (3) holds. 8. Prove lim x 1 x 2 +2 x - 1 x +1 = 1. Proof: To show: for any > 0, there exists δ > 0 such that x 2 + 2 x - 1 x + 1 - 1 < , 0 < | x - 1 | < δ. (5) Since x 2 +2 x - 1 x +1 - 1 = ( x - 1)( x +2) x +1 , we take δ 1 = 1. Then | x +2 | = | x - 1+3 | ≤ | x - 1 | +3 < δ 1 + 3 = 1 + 3 = 4 and | x + 1 | = | x - 1 + 2 | ≥ 2 - | x - 1 | > 2 - δ 1 = 2 - 1 = 1. Therefore, when | x - 1 | < δ 1 , the inequality (5) can be reduced to x 2 + 2 x - 1 x + 1 - 1 < 4 1 | x - 1 | = 4 | x - 1 | < and we solve the second inequality above to get δ 2 = 4 . For the given , we take δ = min { δ 1 , δ 2 } = min { 1 , 4 } , whenever 0 < | x - 1 | < δ , (5) holds.

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• Calculus, Limit, lim g, Point C

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