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# Lecture5 limit

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### Lecture5 limit

1. 1. · · · T HE P RECISE D EFINITION OF A L IMIT · · · (SMS1102 C ALCULUS 1) L05-M041 http://staff.iium.edu.my/suryadi/DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 1 / 15
2. 2. T HE P RECISE D EFINITION OF A L IMIT I NTRODUCTION ”arbitraly” limit ”gets arbitrarily close to” ⇓ ⇓ ”precisely deﬁned (we chose)” ”precise limit”DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 2 / 15
3. 3. T HE P RECISE D EFINITION OF A L IMIT I NTRODUCTIONEXAMPLE 1Consider the function y = 2x − 1 (1)near x0 = 4. Intuitively it is clear that y isclose to 7 when x is close to 4, so lim (2x − 1) = 7 (2) x→4However, how close to x = 4 does x haveto be so that y = 2x − 1 differs from 7 by,say, less than 2 units?SOLUTION 1For what value of x is |y − 7| < 2?. In termof x F IGURE : Keeping x within 1 unit or x0 = 4 will keep y within 2 units |y − 7| = |(2x − 1) − 7| = |2x − 8| (3) of y = 7 |2x − 8| < 2 (4) −1 < x − 4 < 1 (5) DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 3 / 15
4. 4. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT To deﬁne δ > 0 so that keeping x within the interval (x − δ, x + δ) will keep f (x) 1 1 within the interval L − ,L + 10 10DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 4 / 15
5. 5. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT D EFINITION Let f (x) be deﬁned on an open interval about x0 , except possibly at x0 itself. We say that the limit of f (x) as x approaches x0 is the number L, and write lim f (x) = L, (6) x→x0 if, for every number > 0, there exists a corresponding number δ > 0 such that for all x, 0 < |x − x0 | < δ =⇒ |f (x) − L| < . (7) N OTE : 1 When the interval of values δ about x0 is symmetric, we could take δ to be half the length of that interval. 2 When such symmetry is absent, we can take δ to be the distance from x0 to theinterval’s nearer endpoint.DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 5 / 15
6. 6. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT the limit of f (x) as x approaches x0 is the number L, lim f (x) = L x→x0 if, for every number > 0, there exists a corresponding number δ > 0 such that for all x, 0 < |x − x0 | < δ =⇒ |f (x) − L| < .DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 6 / 15
7. 7. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT the limit of f (x) as x approaches x0 is the number L, lim f (x) = L x→x0 if, for every number > 0, there exists a corresponding number δ > 0 such that for all x, 0 < |x − x0 | < δ =⇒ |f (x) − L| < .DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 7 / 15
8. 8. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT the limit of f (x) as x approaches x0 is the number L, lim f (x) = L x→x0 if, for every number > 0, there exists a corresponding number δ > 0 such that for all x, 0 < |x − x0 | < δ =⇒ |f (x) − L| < .DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 8 / 15
9. 9. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMITEXAMPLE 2Show that lim (5x − 3) = 2 x→1SOLUTIONIn deﬁnition of limit: x0 = 1, f (x) = 5x − 3and L = 2, then for ∀ > 0, ∃δ > 0 suchthat ∀x 0 < |x − 1| < δ =⇒ |f (x) − 2| < (8)Fine δ from the −inequality, |(5x − 3) − 2| = |5x − 5| < (9) 5|x − 1| < (10) |x − 1| < /5 (11) F IGURE : If f (x) = 5x − 3, then 0 < |x − 1| < /5, guaranties thatTake δ = /5. Any smaller positive δ will |f (x) − 2| <make 0 < |x − 1| < δ ⇒ |f (x) − 2| < DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 9 / 15
10. 10. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMITEXAMPLE 3Prove the following limits: lim x = x0 (12) lim k = k (13) x→x0 x→x0 DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 10 / 15
11. 11. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMITE XERCISE 2.3Use the graph to ﬁnd a δ > 0 such that for all x, 0 < |x − x0 | < δ =⇒ |f (x) − L| < . DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 11 / 15
12. 12. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMITE XERCISE 2.3Use the graph to ﬁnd a δ > 0 such that for all x, 0 < |x − x0 | < δ =⇒ |f (x) − L| < . DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 12 / 15
13. 13. T HE P RECISE D EFINITION OF A L IMIT F INDING δ A LGEBRAICALLY FOR G IVENH OW TO FIND A LGEBRAICALLY A δ FOR A G IVEN f , L, x0 , ANDThe process to ﬁnding δ > 0 such that for all x 0 < |x − x0 | < δ =⇒ |f (x) − L| < .can be accomplished in two steps: 1 Solve the inequality |f (x) − L| < to ﬁnd an open interval (a, b) containing x0 on which the inequality holds for all x = X0 2 Find a value of δ > 0 that places the open interval (x0 − δ, x0 + δ) centered at x0 inside the interval (a, b). The inequality |f (x) − L| < will hold for all x = x0 in this δ-interval. DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 13 / 15
14. 14. T HE P RECISE D EFINITION OF A L IMIT F INDING δ A LGEBRAICALLY FOR G IVENE XAMPLE 4For the limit √ lim x −1=2 (14) x→5ﬁnd a δ > 0 that work for = 1. That is, ﬁnd a δ > 0 such that for all x, √ 0 < |x − 5| < δ =⇒ | x − 1 − 2| < 1.S OLITION 1 Solve the inequality √ | x − 1 − 2| < 1 to ﬁnd an interval containing x0 = 5 on which the inequality holds for al x = x0 2 Find a value δ > 0 of to place the centered interval 5 − δ < x < 5 + δ (centered at x0 = 5 ) inside the interval (2, 10). DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 14 / 15
15. 15. T HE P RECISE D EFINITION OF A L IMIT F INDING δ A LGEBRAICALLY FOR G IVENE XAMPLE 5Prove that limx→2 = f (x) = 4 x 2, x = 2 f (x) = (15) 1, x = 2S OLITIONTo show that given > 0 there is exist δ > 0 suchthat for all x, 0 < |x − 2| < δ ⇒ |f (x) − 4| < . 1 Solve the inequality |f (x) − 4| < to ﬁnd an interval containing x0 = 2 on which the inequality holds for al x = x0 2 Find a value δ > 0 of to place the centered − interval 2√ δ < x√ 2 + δ inside the < interval 4− , 4+ . DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 15 / 15