Some Trigonometric Limits

Transcription

1 Some Trigonometric Limits Mathematics 11: Lecture 7 Dan Sloughter Furman University September 20, 2007 Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

2 The squeeze theorem Supppose p > 0 and, for all in (a p, a) (a, a + p), we have f () g() h(). If then lim a g() = L. lim f () = L and lim h() = L, a a Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

3 Proof of the squeeze theorem Given ɛ > 0, we may find a δ > 0 such that if 0 < a < δ, then f () L < ɛ and h() L < ɛ. Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

4 Proof of the squeeze theorem Given ɛ > 0, we may find a δ > 0 such that if 0 < a < δ, then f () L < ɛ and h() L < ɛ. Hence if 0 < a < δ, we have L ɛ < f () g() h() < L + ɛ, that is g() L < ɛ. Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

5 Eample ( ) 1 Evaluate lim 2 sin 0 2. Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

6 Eample ( ) 1 Evaluate lim 2 sin 0 2. Note: Since 1 sin ( ) 1 1 for all 0, we have 2 ( ) sin 2 2 for all 0. Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

7 Eample ( ) 1 Evaluate lim 2 sin 0 2. Note: Since 1 sin ( ) 1 1 for all 0, we have 2 ( ) sin 2 2 for all 0. Since lim 2 = 0 = lim ( 2 ), it follows from the squeeze theorem 0 0 that ( ) 1 lim 2 sin 0 2 = 0. Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

8 Eample (cont d) Graphs of y = 2, y = 2, and y = 2 sin ( 1 2 ) : Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

9 Sine at 0 P sin( ) O A B Suppose 0 < < π 2 as in the figure. Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

10 Sine at 0 P sin( ) O A B Suppose 0 < < π 2 as in the figure. Note: 0 < sin() = AP < BP <. Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

11 Sine at 0 P sin( ) O A B Suppose 0 < < π 2 as in the figure. Note: 0 < sin() = AP < BP <. Since lim = 0, it follows that lim sin() = Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

12 Sine at 0 P sin( ) O A B Suppose 0 < < π 2 as in the figure. Note: 0 < sin() = AP < BP <. Since lim = 0, it follows that lim sin() = Similarly, lim sin() = Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

13 Sine at 0 P sin( ) O A B Suppose 0 < < π 2 as in the figure. Note: 0 < sin() = AP < BP <. Since lim = 0, it follows that lim sin() = Similarly, lim sin() = 0. 0 Hence, lim 0 sin() = 0. 0 Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

14 Sine at 0 P sin( ) O A B Suppose 0 < < π 2 as in the figure. Note: 0 < sin() = AP < BP <. Since lim = 0, it follows that lim sin() = Similarly, lim sin() = Hence, lim 0 sin() = 0. Problem 51 asks you to verify that, for any a, lim a sin() = sin(a). Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

15 Cosine at 0 For π 2 < < π 2, cos() = 1 sin 2 (). Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

16 Cosine at 0 For π 2 < < π 2, cos() = 1 sin 2 (). Hence lim cos() = lim 1 sin 2 () = Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

17 Cosine at 0 For π 2 < < π 2, cos() = Hence 1 sin 2 (). lim cos() = lim 1 sin 2 () = Problem 52 asks your to verify that, for any a, lim a cos() = cos(a). Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

18 Two important limits sin() lim = 1 0 Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

19 Two important limits sin() lim = cos() lim = 0 0 Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

20 Proof sin( ) P Q O A B Suppose 0 < < π 2. From the above figure, we see that the area of OBP is less than the area of sector OBP, which is less than the area of OBQ. Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

21 Proof (cont d) That is, 1 2 sin() < 2π π < 1 2 tan(). Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

22 Proof (cont d) That is, Hence 1 < 1 2 sin() < 2π π < 1 2 tan(). sin() < 1 cos(). Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

23 Proof (cont d) That is, Hence 1 < 1 2 sin() < 2π π < 1 2 tan(). sin() < 1 cos(). Then cos() < sin() < 1. Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

24 Proof (cont d) That is, Hence 1 < 1 2 sin() < 2π π < 1 2 tan(). sin() < 1 cos(). Then cos() < sin() < 1. Note: this inequality also holds for π 2 < < 0 since cos( ) = cos() and sin( ) = sin() = sin(). Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

25 Proof (cont d) Using the squeeze theorem, and the fact that lim cos() = 1, it 0 follows that sin() lim = 1. 0 Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

26 Proof (cont d) The second limit now follows from the first: ( ) ( ) 1 cos() 1 cos() 1 + cos() lim = lim cos() 1 cos 2 () = lim 0 (1 + cos()) sin 2 () = lim 0 (1 + cos()) ( ) ( ) sin() sin() = lim cos() = = 0. Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

27 Eample We may use the previous results to evaluate other limits: sin(3) 3 sin(3) lim = lim sin(3) = 3 lim 0 3 = (3)(1) = 3. Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14

28 Eample Using the idea of the previous eample: sin(4) lim 0 sin(3) = lim 0 = 4 3 lim 0 = = 4 3. sin(4) sin(3) sin(4) 4 sin(3) 3 Dan Sloughter (Furman University) Some Trigonometric Limits September 20, / 14