Upcoming SlideShare
×

# Lesson 2: Limits and Limit Laws

7,749 views

Published on

The concept of limit formalizes the notion of closeness of the function values to a certain value "near" a certain point. Limits behave well with respect to arithmetic--usually. Division by zero is always a problem, and we can't make conclusions about nonexistent limits!

Published in: Technology, Education
2 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total views
7,749
On SlideShare
0
From Embeds
0
Number of Embeds
260
Actions
Shares
0
240
0
Likes
2
Embeds 0
No embeds

No notes for slide

### Lesson 2: Limits and Limit Laws

1. 1. Section 2.2–3 The Concept of Limit Limit Laws Math 1a February 4, 2008 Announcements Syllabus available on course website Homework for Wednesday 2/6: Practice 2.2: 1, 3, 5, 7, 13, 15; 2.3: 1, 3, 7, 13, 15, 17 Turn-in 2:2: 2, 4, 6, 8; 2.3: 2, 20, 38 Homework for Monday 2/11: 2.2.28, 2.3.30, 2.4.34 ALEKS due Wednesday 2/20
2. 2. Outline The Concept of Limit Heuristics Errors and tolerances Pathologies Limit Laws Easy laws The direct substitution property Limits by algebra Two more limit theorems
3. 3. Zeno’s Paradox That which is in locomotion must arrive at the half-way stage before it arrives at the goal. (Aristotle Physics VI:9, 239b10)
4. 4. Heuristic Deﬁnition of a Limit Deﬁnition We write lim f (x) = L x→a and say “the limit of f (x), as x approaches a, equals L” if we can make the values of f (x) arbitrarily close to L (as close to L as we like) by taking x to be suﬃciently close to a (on either side of a) but not equal to a.
5. 5. The error-tolerance game L a
6. 6. The error-tolerance game L a
7. 7. The error-tolerance game L a
8. 8. The error-tolerance game This tolerance is too big L a
9. 9. The error-tolerance game L a
10. 10. The error-tolerance game Still too big L a
11. 11. The error-tolerance game L a
12. 12. The error-tolerance game This looks good L a
13. 13. The error-tolerance game So does this L a
14. 14. Examples Example Find lim x 2 if it exists. x→0
15. 15. Examples Example Find lim x 2 if it exists. x→0 Example |x| Find lim if it exists. x→0 x
16. 16. Examples Example Find lim x 2 if it exists. x→0 Example |x| Find lim if it exists. x→0 x Example 1 Find lim+ if it exists. x→0 x
17. 17. Examples Example Find lim x 2 if it exists. x→0 Example |x| Find lim if it exists. x→0 x Example 1 Find lim+ if it exists. x→0 x Example π Find lim sin if it exists. x→0 x
18. 18. What could go wrong? How could a function fail to have a limit? Some possibilities: left- and right- hand limits exist but are not equal The function is unbounded near a Oscillation with increasingly high frequency near a
19. 19. Precise Deﬁnition of a Limit Let f be a function deﬁned on an some open interval that contains the number a, except possibly at a itself. Then we say that the limit of f (x) as x approaches a is L, and we write lim f (x) = L, x→a if for every ε > 0 there is a corresponding δ > 0 such that if 0 < |x − a| < δ, then |f (x) − L| < ε.
20. 20. The error-tolerance game = ε, δ L a
21. 21. The error-tolerance game = ε, δ L+ε L L−ε a
22. 22. The error-tolerance game = ε, δ L+ε L L−ε a − δaa + δ
23. 23. The error-tolerance game = ε, δ This δ is too big L+ε L L−ε a − δaa + δ
24. 24. The error-tolerance game = ε, δ L+ε L L−ε a −aδ δ a+
25. 25. The error-tolerance game = ε, δ This δ looks good L+ε L L−ε a −aδ δ a+
26. 26. The error-tolerance game = ε, δ So does this δ L+ε L L−ε aa a δ δ − +
27. 27. Meet the Mathematician: Augustin Louis Cauchy French, 1789–1857 Royalist and Catholic made contributions in geometry, calculus, complex analysis, number theory created the deﬁnition of limit we use today but didn’t understand it
28. 28. Outline The Concept of Limit Heuristics Errors and tolerances Pathologies Limit Laws Easy laws The direct substitution property Limits by algebra Two more limit theorems
29. 29. Limit Laws Suppose that c is a constant and the limits lim f (x) and lim g (x) x→a x→a exist. Then 1. lim [f (x) + g (x)] = lim f (x) + lim g (x) x→a x→a x→a 2. lim [f (x) − g (x)] = lim f (x) − lim g (x) x→a x→a x→a 3. lim [cf (x)] = c lim f (x) x→a x→a 4. lim [f (x)g (x)] = lim f (x) · lim g (x) x→a x→a x→a
30. 30. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a
31. 31. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) x→a x→a
32. 32. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a
33. 33. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a 7. lim c = c x→a 8. lim x = a x→a
34. 34. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a 7. lim c = c x→a 8. lim x = a x→a 9. lim x n = an x→a √ √ 10. lim n x = n a x→a
35. 35. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a 7. lim c = c x→a 8. lim x = a x→a 9. lim x n = an (follows from 6 and 8) x→a √ √ 10. lim n x = n a x→a
36. 36. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a 7. lim c = c x→a 8. lim x = a x→a 9. lim x n = an (follows from 6 and 8) x→a √ √ 10. lim n x = n a x→a n 11. lim f (x) = n lim f (x) (If n is even, we must additionally x→a x→a assume that lim f (x) > 0) x→a
37. 37. Direct Substitution Property Theorem (The Direct Substitution Property) If f is a polynomial or a rational function and a is in the domain of f , then lim f (x) = f (a) x→a
38. 38. Limits do not see the point! (in a good way) Theorem If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L. x→a x→a
39. 39. Limits do not see the point! (in a good way) Theorem If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L. x→a x→a Example x 2 + 2x + 1 Find lim , if it exists. x→−1 x +1
40. 40. Limits do not see the point! (in a good way) Theorem If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L. x→a x→a Example x 2 + 2x + 1 Find lim , if it exists. x→−1 x +1 Solution x 2 + 2x + 1 Since = x + 1 whenever x = −1, and since x +1 x 2 + 2x + 1 lim x + 1 = 0, we have lim = 0. x→−1 x→−1 x +1
41. 41. Finding limits by algebraic manipulations Example √ x −2 Find lim . x→4 x −4
42. 42. Finding limits by algebraic manipulations Example √ x −2 Find lim . x→4 x −4 Solution √ √ √ 2 Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2).
43. 43. Finding limits by algebraic manipulations Example √ x −2 Find lim . x→4 x −4 Solution √ 2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). So √ √ x −2 x −2 lim = lim √ √ x→2 x − 4 x→2 ( x − 2)( x + 2) 1 1 = lim √ = x→2 x +2 4
44. 44. Finding limits by algebraic manipulations Example √ x −2 Find lim . x→4 x −4 Solution √ 2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). So √ √ x −2 x −2 lim = lim √ √ x→2 x − 4 x→2 ( x − 2)( x + 2) 1 1 = lim √ = x→2 x +2 4 Example√ √ 3 x− 3a Try lim . x→a x −a
45. 45. Two More Important Limit Theorems Theorem If f (x) ≤ g (x) when x is near a (except possibly at a), then lim f (x) ≤ lim g (x) x→a x→a (as usual, provided these limits exist). Theorem (The Squeeze/Sandwich/Pinching Theorem) If f (x) ≤ g (x) ≤ h(x) when x is near a (as usual, except possibly at a), and lim f (x) = lim h(x) = L, x→a x→a then lim g (x) = L. x→a
46. 46. We can use the Squeeze Theorem to make complicated limits simple.
47. 47. We can use the Squeeze Theorem to make complicated limits simple. Example 1 Show that lim x 2 sin = 0. x→0 x
48. 48. We can use the Squeeze Theorem to make complicated limits simple. Example 1 Show that lim x 2 sin = 0. x→0 x Solution We have for all x, 1 −x 2 ≤ x 2 sin ≤ x2 x The left and right sides go to zero as x → 0.