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- 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. 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. 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. 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. The error-tolerance game L a
- 6. The error-tolerance game L a
- 7. The error-tolerance game L a
- 8. The error-tolerance game This tolerance is too big L a
- 9. The error-tolerance game L a
- 10. The error-tolerance game Still too big L a
- 11. The error-tolerance game L a
- 12. The error-tolerance game This looks good L a
- 13. The error-tolerance game So does this L a
- 14. Examples Example Find lim x 2 if it exists. x→0
- 15. Examples Example Find lim x 2 if it exists. x→0 Example |x| Find lim if it exists. x→0 x
- 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. 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. 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. 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. The error-tolerance game = ε, δ L a
- 21. The error-tolerance game = ε, δ L+ε L L−ε a
- 22. The error-tolerance game = ε, δ L+ε L L−ε a − δaa + δ
- 23. The error-tolerance game = ε, δ This δ is too big L+ε L L−ε a − δaa + δ
- 24. The error-tolerance game = ε, δ L+ε L L−ε a −aδ δ a+
- 25. The error-tolerance game = ε, δ This δ looks good L+ε L L−ε a −aδ δ a+
- 26. The error-tolerance game = ε, δ So does this δ L+ε L L−ε aa a δ δ − +
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Finding limits by algebraic manipulations Example √ x −2 Find lim . x→4 x −4
- 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. 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. 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. 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. We can use the Squeeze Theorem to make complicated limits simple.
- 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. 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.

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