## anonymous 5 years ago By the method of delta-epsilon prove limx->5 ((1/2)x+6)= 8 and a1/2

1. amistre64

ugh.... epsilons and deltas.....

2. amistre64

f(x)-L < e |x-c|<d

3. amistre64

at least its a line and not a curve

4. anonymous

i know all of that sorry let me tell you where am having a problem

5. anonymous

i know all of that sorry let me tell you where am having a problem

6. anonymous

ok i simplified it to get 1/2 (x-2) < e so how do i find my delta from there?

7. amistre64

|x-5| < d right? id have to work it out meself to recall how to do it lol

8. anonymous

yes

9. anonymous

yes

10. anonymous

its |x-5|<d

11. amistre64

a1/2 ?? whats that mean? -e < (1/2)x+6) - 8 < e

12. anonymous

8 and a half

13. amistre64

-e < (1/2)x-2 < e 2 -e < (1/2)x < 2+e 2(2 -e) < x < 2(2+e)

14. amistre64

0 < x-5 <d 5 < x < 5+d

15. amistre64

5+d < 2(2+e) i got know idea if this is working lol

16. anonymous

arent we suppose to be converting x-2 to x-5?..no?

17. anonymous

usually you work these problems backwards.

18. anonymous

this one is tricky...

19. anonymous

start with | $|.5x+6-8.5|<\epsilon$

20. anonymous

yes

21. anonymous

$-\epsilon < .5x-2.5< \epsilon$

22. amistre64

ohh....8' 1/2 ....

23. anonymous

$2.5-\epsilon<.5x<2.5+\epsilon$

24. anonymous

$5-2\epsilon<x<5+2\epsilon$

25. amistre64

looks familiar so far lol

26. anonymous

can u do it in terms of epsilon and delta?

27. anonymous

can u do it in terms of epsilon and delta?

28. anonymous

how am i doing so far?

29. anonymous

good

30. amistre64

well, other than having the right numbers; looks the same as mine lol

31. anonymous

oh my delta will depend on epsilon. i am just trying to figure out what the delta needs tgo be

32. anonymous

kk

33. anonymous

the game is played like this: i say the limit is L. you say ok, make it within epsilon of L. i say ok use this delta which obviously depends on epsilon. that is delta will be written in terms of epsilon

34. anonymous

ok but in class we normally would deleta write where u wrote your epsilon but its problem a different example

35. anonymous

wait

36. anonymous

how'd you get the .5 to 5?

37. anonymous

i am working the problem backwards. then the proof is to work it forwards. you say for example pick delta such that $5-2\epsilon < \delta< 5+2\epsilon$

38. anonymous

the 5?

39. anonymous

we had $2.5- \epsilon<.5x<2.5+\epsilon$

40. anonymous

to get x by itself i doubled everything

41. anonymous

ok so now we have $5-2\epsilon <x < 5+2\epsilon$

42. anonymous

this says the same thing as $|x-5|<2\epsilon$

43. anonymous

so the delta i pick is $2\epsilon$

44. anonymous

oh ok sorry i had to move from the pc a sec

45. anonymous

because that is what we wanted. we wanted to say that we could find a delta such that if $|x-5|<\delta$ then $|.5x-6+8.5|<\epsilon$

46. anonymous

don't forget what the goal is. you want to show that $lim_{x->5} (.5x+6)=8$

47. anonymous

8 1/2

48. anonymous

meaning that given any epsilon > 0 y ou can find a delta such that if $|x-5|<\delta$ then $|.5x+6-8.5|<\epsilon$

49. anonymous

ok truthfully i dont understand what u did

50. anonymous

i say pick $\delta = 2 \epsilon$ i win by running the algebra backwards

51. anonymous

the way you have to do these problem is work backwards. you have to find the delta, which depends on epsilon. i.e. $\delta = delta(\epsilon)$

52. anonymous

delta will be an expression in epsilon. so you start with the thing that you want, namely $|.5x+6-8.5|<\epsilon$ do some algebra, and see what delta you get.

53. anonymous

i got $\delta = 2\epsilon$

54. anonymous

ok

55. anonymous

thank you!

56. anonymous

run the algebra forward and see that it works. just take the steps i wrote and write them backwards. in other words start with $|x-5|<2\delta$ and see that you get $|2.x+6-8.5|<\epsilon$

57. anonymous

typo

58. anonymous

i meant start with $|x-5|<2\epsilon$

59. anonymous

i will write it if you like

60. anonymous

61. anonymous

let me write it and maybe it will make sense. hold on and i will write it all out.

62. anonymous

okat

63. anonymous

You want to show that given any $\epsilon>0$ you can find a $\delta$ such that if $|x-5|<\delta$ then $|.5x+6-8.5|<\epsilon$ let $\delta=2\epsilon$ then $|x-5|<2\epsilon$ means $-2\epsilon<x-5<\epsilon$ giving $5-2\epsilon < x < 5+2\epsilon$ gives $2.5-\epsilon < .5x < 2.5 +\epsilon$ so $-\epsilon < .5x-2.5<\epsilon$ or $|.5x-2.5|<\epsilon$

64. anonymous

sorry it took so long to write. but i never would have found delta = 2 epsilon without working the problem backwards

65. anonymous

hope this helps

66. anonymous

wow THANK YOU SO MUCH

67. anonymous

welcome.

68. anonymous

ok like thank you so much i get it but i never would have figured that out on my own thanks a billion seriously

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