## anonymous 3 years ago Use numerical and graphical evidence to conjecture value(s) for the limits, if they exist. If not, describe what is happening at x = a graphically.

1. anonymous

1) lim x-> 1 (x^2 - 1)/(x + 2) 2) lim x-> 1 (x^2 + 1)/(x - 1)

2. anonymous

@UnkleRhaukus umm can u help?

3. UnkleRhaukus

1) $\lim\limits_ {x\to1}\frac{ x^2 - 1}{x + 2}$ what happens when we put x=1 into the fraction

4. anonymous

hmm its becomes 0/3?

5. UnkleRhaukus

yeah, do you know what that simplifies to ?

6. anonymous

umm 0?

7. UnkleRhaukus

correct,

8. anonymous

okay so how would i describe it?

9. UnkleRhaukus

can you fill in the table |dw:1359329822568:dw|

10. anonymous

yes i can

11. anonymous

okay let me try this

12. anonymous

so for the -2 one i'm dividing -5 to 0 so will it be -5/0 since i cant divide that?

13. anonymous

ok so i got |dw:1359330562610:dw|

14. anonymous

so did i get it?

15. UnkleRhaukus

|dw:1359330502446:dw|

16. anonymous

are u there?

17. anonymous

ok

18. anonymous

19. anonymous

umm are u there?

20. UnkleRhaukus

(sorry my internet cut out for a moment there) um you got most of them right the first time but check the case when x=-1 agian |dw:1359331049113:dw|

21. anonymous

okay

22. anonymous

so then i draw the graph? right ?

23. UnkleRhaukus

yeah not plot those points |dw:1359331297137:dw|

24. UnkleRhaukus

not*now

25. anonymous

ok

26. UnkleRhaukus

|dw:1359331448238:dw| maybe i should have chosen some better x points

27. anonymous

what are the question marks and the lines?

28. UnkleRhaukus

ive put question marks for for the x=-2, y=-5/0 bit

29. anonymous

ohh okay

30. UnkleRhaukus

|dw:1359331760380:dw|

31. anonymous

so its a curve?

32. UnkleRhaukus

if i chose better x values we would have got something like this

33. anonymous

okay

34. anonymous

so that will be the end of the problem?

35. UnkleRhaukus

|dw:1359332271558:dw|

36. UnkleRhaukus

|dw:1359332425565:dw|

37. UnkleRhaukus

|dw:1359332578040:dw|

38. anonymous

ok so this is the final graph?

39. UnkleRhaukus

this shape is called a hyperbola at x=-2 y is undefined, this is because if we look at the graph is should be +∞ and also -∞  but it can't be more than one point we call this a vertical Asymptote

40. anonymous

okay

41. UnkleRhaukus

|dw:1359332784659:dw|

42. anonymous

ohh ok i get it

43. UnkleRhaukus

to describe what is happening at x=a if $$a < -2$$ the value will be on the bottom curve if $$a = -2$$ the value is undefined if $$a >2$$ the value will be on the top curve

44. UnkleRhaukus

__________ for 2) $\lim\limits_{x\to 1} \frac{x^2 + 1}{x - 1}$ the value is undefined as we are trying to taking the limit at the asymptote

45. anonymous

okay

46. UnkleRhaukus

i hope i haven't confused you too much

47. anonymous

noo u havent so i'll do the same this with lim x-> 1 (x^2 + 1)/(x -1)

48. UnkleRhaukus

another way of saying the value is undefined is to say the limit does not exist

49. anonymous

okay

50. anonymous

so i'll do the same for lim x-> 1 (x^2 + 1)/(x -1)?

51. UnkleRhaukus

yeah, i would use x =-5,-4,-3,-2,-1,0,2,3,4,5 this time

52. anonymous

ok thanks!! :)

53. UnkleRhaukus

ok im off, good luck

54. anonymous

hey so i did it is this correct? |dw:1359335654116:dw| |dw:1359335759751:dw|

55. anonymous

@satellite73 can u help with this?

56. anonymous

@zepdrix ??

57. anonymous

@jim_thompson5910 ??