## anonymous 4 years ago what is this limit $\lim_{x \rightarrow \pi/2}(xcosx)$

1. anonymous

$\lim_{x \rightarrow \pi/2}(xcosx)$

2. Mertsj

0

3. anonymous

mertsj do you mind demonstrating that please?

4. Mertsj

pi/2times cospi/2=pi/2times 0 = 0

5. anonymous

ok thanks would you mind helping me with this one also

6. anonymous

$\lim_{x \rightarrow -7}(x^2 + 10x + 21/x + 7)$ i've got -98/0. i dont know what to do from there

7. Mertsj

You can't divide by 0 so divide numerator so another approach is needed.

8. Mertsj

Factor and reduce first

9. anonymous

ok i was thinking along the lines as approach -7 from left and right

10. anonymous

(x+7)(x+3)/(x+7)...so the x + 7 cancels and im left with x + 3

11. Mertsj

yes

12. Mertsj

Now replace the x with -7

13. anonymous

i see.

14. anonymous

i have this one that is somewhat similar but you end up with 0/0 $\lim_{x \rightarrow 1}(x^2 - 4x + 3/ x^2 + 2x -3)$

15. anonymous

same approach again right? factor and reduce?

16. Mertsj

yes

17. anonymous

(x-3)(x-1)/ (x-3)(x+1)...?

18. anonymous

is that the right structure

19. Mertsj

|dw:1326943979278:dw|

20. anonymous

oh wow im getting this thanks for the help. would you be willing to continue working with me? i have about 8 more

21. Mertsj

I am willing but I might be a little distracted as my daughter is here and we are visiting.

22. anonymous

ok no problem

23. anonymous

$\lim_{x \rightarrow -10}(x^2 - 17x + 70/ x + 10)$

24. anonymous

when i attempt this...

25. anonymous

i get 140/0...so i take the factor and reduce approach

26. anonymous

and..

27. Mertsj

Yes factor and reduce

28. Mertsj

|dw:1326944396601:dw|

29. anonymous

(x+10)(x+7)/ (x+10)....x+7...-10+7= -3?

30. anonymous

i keep getting my signs wrong when i factor

31. Mertsj

It won't reduce. Did you copy it right?

32. anonymous

why couldnt it have been (x+10)(x+7)...it will also give you 70 right

33. Mertsj

Middle term is negative

34. anonymous

yeah thats right i remember now

35. Mertsj

Did you copy it right?

36. anonymous

what do you mean

37. anonymous

it does not exist

38. anonymous

i see that

39. Mertsj

Perhaps you wrote the problem down incorrectly

40. anonymous

yeah now i understand

41. anonymous

i have this next one...

42. Mertsj

Did you get that last one?

43. anonymous

yeah

44. Mertsj

What did you get?

45. anonymous

$\lim_{x \rightarrow 2}(x^2 - 4/2x^2 -5x + 2)$

46. anonymous

I got DNE (does not exist) my software accepted the answer

47. anonymous

how exactly would you factor the denominator in the previous problem. it is set up strangely for me

48. Mertsj

That one should factor and reduce.

49. Mertsj

(2x-1)(x-2)

50. anonymous

ok i was confused because i thought somehow it must add up to -5

51. Mertsj

So you end up with (x+2)/(2x-1)

52. anonymous

gotcha

53. anonymous

next is...

54. anonymous

$\lim_{x \rightarrow 4}(\sqrt{b+0}-2/b-4)$

55. anonymous

0/0

56. anonymous

do you still factor here? doesnt look like it

57. Mertsj

It won't factor but why does it have sqrt(b+0) since that is just sqrt(b)?

58. anonymous

its sqrt (b +0)...i would guess you do 4 +0 then squared root your answer then subtract 2

59. Mertsj

But the denominator will be 0 and that is undefined.

60. anonymous

yeah i got 0/0 i was thinking theres something we could do next

61. Mertsj

multiply the numerator and denominator by (sqrt(b)+2)

62. anonymous

if the numerator and denominator are both zeros. wouldnt mulitplying bring us back to square 1 with 0/0

63. Mertsj

|dw:1326945956921:dw|

64. anonymous

i see

65. anonymous

|dw:1326946154087:dw|

66. anonymous

the limit is as x approaches -9

67. Mertsj

Multiply numerator and denominator by 9x

68. Mertsj

|dw:1326946478456:dw|

69. Mertsj

lim as x approaches -9 is -1/81

70. anonymous

thanks im almost through

71. anonymous

Let 11x−36≤f(x)≤x2+5x−27 . Use the Squeeze Theorem to determine limx→3f(x) =

72. Mertsj

ok but I still want to revisit that DNE one

73. anonymous

ok when we finish we can go back

74. anonymous

this one that i posted is a new concept so its unlike what we have been practicing so im clueless

75. Mertsj

I see that. I have never heard of the squeeze theorem but I would assume that we should evaluate both expressions when x is 3

76. anonymous

i got 33 and -3 respectively. what do i do with these two numbers to come up with a final answer

77. Mertsj

You should get -3 for both of them and that is the limit. I just read it on the internet.

78. Mertsj

redo the one you got the 33 for

79. anonymous

yeah you're right i entered it incorrectly in my calculator

80. anonymous

|dw:1326947257959:dw| as x approaches 0

81. Mertsj

82. anonymous

ok

83. Mertsj

Have you tried typing it into wolframalpha?

84. TuringTest

L'hospital does this one easily

85. anonymous

|dw:1326947625929:dw|

86. Mertsj

Well let's see it.

87. anonymous

as x approaches 0

88. Mertsj

I'll turn you over to Turing now.

89. anonymous

thanks for all the help Mertsj I really appreciate it

90. Mertsj

yw

91. Mertsj

92. anonymous

ok I hope to see you around again. Enjoy

93. TuringTest

@mertsj, yes, well, done @Jinnie: do you know L'hospital's rule? derivative of top and bottom gives $\frac{9\cos(9x)}{1}$

94. Mertsj

Thanks

95. anonymous

hey turningtest thats my first time hearing about that rule

96. TuringTest

can you take the derivative of sin(9x) ?

97. anonymous

no i dont know how

98. TuringTest

Then you can't use L'hospital, which unfortunately means I can't help you as well as I hoped. I can tell you that $\lim_{x \rightarrow 0}\frac{\sin x}{x}=1$but proof of that are bit complicated without L'hospital's rule. All I can suggest is to try to use that fact to help you. anyway the answer is that the limit in your question is 9

99. anonymous

100. TuringTest

have you learned derivatives yet at all?

101. anonymous

nah we just started school last week

102. anonymous

|dw:1326948473758:dw| lim as x approaches 0. this is all i need help with

103. TuringTest

then all I can do is give answers without explanation, because the way I solve these is with a method you don't know. If you just want the answer I refer you to wolfram alpha as mertsj did. Here is your question posted there: http://www.wolframalpha.com/input/?i=limit+x+to+0+sin%289x%29%2Fx learn to use this site, but it will not help you understand, so don't rely on it.

104. TuringTest

this last one is 1/3 try typing it into wolfram to see if I'm right

105. anonymous

alright thanks for the help

106. anonymous

im done thanks to all

107. anonymous

good night hope to see everyone again

108. TuringTest

goodnight :)