## roselin whats the lim x goes to infinity of 2/square root of x one year ago one year ago

1. JoãoVitorMC

0

2. roselin

yes

3. zepdrix

Rose if you're not convinced, just plug a large number into your calculator to check! :) if you plug one bajillion in for x, what value do you get for the limit? $\frac{ 2 }{ \sqrt{bajillion} }\approx \frac{ 2 }{ 999999999 } \approx 0$

4. JoãoVitorMC

yeah, but do what zepdrix said only for check, because if you do this in a test for example, it will be wrong because infinity is NOT a number...

5. roselin

okay

6. roselin

it is always 0 ?

7. JoãoVitorMC

no! it will depends on what you are taking the limit, always try to simplify the limit, try to do not have a denominator, in this case we can not simplify so we do this

8. roselin

i am trying to solve this problem where x goes to infinity 2+square root of x/ 2- square root of x

9. roselin

i am getting the answer as 1, is it right?

10. JoãoVitorMC

$\lim_{x \rightarrow \infty}\frac{2+\sqrt{x}}{2-\sqrt{x}}$

11. JoãoVitorMC

this one?

12. roselin

yes,

13. JoãoVitorMC

the limit is equal to -1

14. JoãoVitorMC

have you already study L'hospital rule ?

15. roselin

can you please explain how you got -1?

16. roselin

No,

17. JoãoVitorMC

ok let me explain

18. JoãoVitorMC

|dw:1350703643878:dw|

19. JoãoVitorMC

got it?

20. roselin

what did you do with the whole numerator thing? u ended up with X

21. JoãoVitorMC

when x goes to infinity, the other terms with lower power we take them off, and stay only with the higher terms them simply if possible and aply the limit

22. JoãoVitorMC

i will post how i got the numerator

23. roselin

ohh , u mean the limit of 4 and 4 sqrt x = 0?

24. JoãoVitorMC

|dw:1350704041192:dw|

25. JoãoVitorMC

no this i simply take them out because comparing with infinity they make no difference

26. roselin

oh okay,

27. JoãoVitorMC

when you deal with limits on the infinity you can make this assumptions

28. JoãoVitorMC

but take care...

29. JoãoVitorMC

when you see L'hostpital's Rule this limits will became very easy to solve.... you will see

30. roselin

we have not gotten to that yet.

31. JoãoVitorMC

yeah

32. roselin

if you do not mind then, I have one more problem. would you explain that one?

33. JoãoVitorMC

sure

34. roselin

thank you

35. roselin

lim sec x x goes to( -pi/2)^+

36. JoãoVitorMC

let me solve here one moment

37. roselin

okay

38. JoãoVitorMC

it is equal to infinity, i will draw here

39. roselin

okay

40. JoãoVitorMC

|dw:1350704691784:dw|

41. JoãoVitorMC

is it clear?

42. roselin

yes,viewing it.

43. JoãoVitorMC

see, when we have something divide by a number really really really small them it becomes a really really really large number so the limit tends to the infinity

44. roselin

okay, but how did you get that 10?

45. JoãoVitorMC

it is just a notation for this: 0.00000000000000000000000...00000000000001, then we represent this really small or large number with 10 power

46. roselin

oh ,okay. got it.

47. roselin

wait,wait

48. roselin

that (-pi/2)^+ is the same as (-pi/2)?

49. JoãoVitorMC

ALMOST the same, when we have this + is a little more than pi/2

50. JoãoVitorMC

-pi/2, sorry

51. roselin

so, the asnwer will be the same one right?

52. roselin

53. JoãoVitorMC

no, because if we take 1/0 it is not allowed. But 1/0.00000000001 is allowed

54. JoãoVitorMC

we can not divide by 0, NEVER

55. roselin

yes,that is true.

56. JoãoVitorMC

so you got it?

57. roselin

yeah.

58. roselin

Thank you soooo much.

59. JoãoVitorMC

yw :)

60. roselin

good night, byee

61. JoãoVitorMC

good, bye