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abannavong Group Title

I need AP Calculus help!!

  • 2 years ago
  • 2 years ago

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  1. abannavong Group Title
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    i will post the assignment soon!

    • 2 years ago
  2. abannavong Group Title
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    Help with all plz!

    • 2 years ago
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  3. abannavong Group Title
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    • 2 years ago
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  4. jldstuff393 Group Title
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    hi

    • 2 years ago
  5. abannavong Group Title
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    the circled questions btw

    • 2 years ago
  6. jldstuff393 Group Title
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    for the first one, its just 8 thats posted?

    • 2 years ago
  7. jldstuff393 Group Title
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    i mean circled

    • 2 years ago
  8. abannavong Group Title
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    its numbers 10, 14, 16, 18, 22

    • 2 years ago
  9. jldstuff393 Group Title
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    oh i was looking at the first one

    • 2 years ago
  10. abannavong Group Title
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    kk

    • 2 years ago
  11. jldstuff393 Group Title
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    oh man, that second one's really hard to read lol so lets do the first one

    • 2 years ago
  12. abannavong Group Title
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    kk

    • 2 years ago
  13. jldstuff393 Group Title
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    k so for 8 what is x approaching? i cant see

    • 2 years ago
  14. abannavong Group Title
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    we dont have to do number 8 lol

    • 2 years ago
  15. jldstuff393 Group Title
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    :\ it looked circled to me lol

    • 2 years ago
  16. abannavong Group Title
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    its only numbers 10, 14, 16, 18, 22

    • 2 years ago
  17. jldstuff393 Group Title
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    do you not have to do the problem in your first attachment?

    • 2 years ago
  18. abannavong Group Title
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    oh yeah duh i forgot i do lolz

    • 2 years ago
  19. jldstuff393 Group Title
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    haha ok cool well i basically cant read any of them cuz they're blurry so can you just type out one that you need help with?

    • 2 years ago
  20. abannavong Group Title
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    kk

    • 2 years ago
  21. abannavong Group Title
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    hang on

    • 2 years ago
  22. abannavong Group Title
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    number 8 its as x goes to infinity

    • 2 years ago
  23. jldstuff393 Group Title
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    so it's as x-> infinity of (2x^2)/(x^2 +3) ? is that right?

    • 2 years ago
  24. abannavong Group Title
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    2x^5/ x^5+3

    • 2 years ago
  25. jldstuff393 Group Title
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    oh ok.

    • 2 years ago
  26. jldstuff393 Group Title
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    have y'all talked about L'Hospital's Rule yet?

    • 2 years ago
  27. abannavong Group Title
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    i dont think so

    • 2 years ago
  28. jldstuff393 Group Title
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    k.

    • 2 years ago
  29. abannavong Group Title
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    what is the L'hospitals rule? can u plz explain

    • 2 years ago
  30. jldstuff393 Group Title
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    i don't think you're supposed to use it if you haven't covered it yet. you don't need it anyway.

    • 2 years ago
  31. jldstuff393 Group Title
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    so you start by considering the terms with the highest exponents

    • 2 years ago
  32. jldstuff393 Group Title
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    if I remember correctly, since x is approaching infinity and it's a rational function, you can rewrite it only including the x's with the greatest exponents

    • 2 years ago
  33. jldstuff393 Group Title
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    so what would you have if you did that?

    • 2 years ago
  34. abannavong Group Title
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    hang on im kinda confused

    • 2 years ago
  35. jldstuff393 Group Title
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    that's ok. so what i'm saying is your next step is to rewrite the limit excluding the +3 in the denominator

    • 2 years ago
  36. abannavong Group Title
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    to exclude the +3 in the denominator how do u do tht?

    • 2 years ago
  37. jldstuff393 Group Title
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    you can do this because when x goes to infinity, 2x^5 and x^5 approach infinity and the 3 basically doesn't change the fraction anymore

    • 2 years ago
  38. jldstuff393 Group Title
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    because if you imagine really high numbers, say x=100, x^5 would be 10000000000 and 2x^5 would be 20000000000 and the +3 in the denominator makes less and less of a difference in the fraction. (10000000000/20000000003 is really close to 10000000000/20000000000)

    • 2 years ago
  39. abannavong Group Title
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    oh ok

    • 2 years ago
  40. jldstuff393 Group Title
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    in other words, the fraction approaches 1/2

    • 2 years ago
  41. jldstuff393 Group Title
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    im sorry, it approaches 2

    • 2 years ago
  42. jldstuff393 Group Title
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    because 2x^5 is in the numerator

    • 2 years ago
  43. abannavong Group Title
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    oh ok so ur limit is basically 2 then right?

    • 2 years ago
  44. jldstuff393 Group Title
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    yup

    • 2 years ago
  45. abannavong Group Title
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    oh ok now i get number 8

    • 2 years ago
  46. jldstuff393 Group Title
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    so in questions like that, where x is approaching infinity, you can disregard any terms that arent the variables with the highest exponent. like if you had an x^4 being added/subtracted somewhere in the fraction, you could disregard that, too.

    • 2 years ago
  47. abannavong Group Title
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    why can u?

    • 2 years ago
  48. jldstuff393 Group Title
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    for the same reason, because as x approaches infinity, they change the fraction less and less, even though it has x in it....those terms approach infinity slower than terms of higher powers.

    • 2 years ago
  49. abannavong Group Title
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    im kinda confused on tht part

    • 2 years ago
  50. jldstuff393 Group Title
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    it's kind of counter-intuitive at first

    • 2 years ago
  51. abannavong Group Title
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    oh ok

    • 2 years ago
  52. jldstuff393 Group Title
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    as x gets greater, the difference between x^5 and x^4 also gets greater. if x is 10, then x^5 is 100000 and x^4 is 10000, x^5 is 10 times greater than x^4. but when x= 100, x^5 becomes 100 times bigger than x^4, and so on, until x^5 is infinitely bigger than x^4 and you can disregard x^4.

    • 2 years ago
  53. abannavong Group Title
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    oh ok now i get it

    • 2 years ago
  54. jldstuff393 Group Title
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    :) cool

    • 2 years ago
  55. abannavong Group Title
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    can u help me with the continuity limits and one-sided ones those ones are the ones that r confusing

    • 2 years ago
  56. jldstuff393 Group Title
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    i can help w one more then i need to get my own work done haha

    • 2 years ago
  57. abannavong Group Title
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    kk

    • 2 years ago
  58. jldstuff393 Group Title
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    so i guess pick one for us to do

    • 2 years ago
  59. abannavong Group Title
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    kk

    • 2 years ago
  60. abannavong Group Title
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    i think i get the rest on my THQ but i need help with the continuity ones

    • 2 years ago
  61. jldstuff393 Group Title
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    k

    • 2 years ago
  62. abannavong Group Title
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    let me post the problem

    • 2 years ago
  63. jldstuff393 Group Title
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    kk

    • 2 years ago
  64. abannavong Group Title
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    \[\lim_{x \rightarrow 4-} \frac{ \sqrt{x}-2 }{ x-4 } \]

    • 2 years ago
  65. jldstuff393 Group Title
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    so first notice that when x = 4, f(x) = 0/0, which is undefined.

    • 2 years ago
  66. abannavong Group Title
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    yeah

    • 2 years ago
  67. jldstuff393 Group Title
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    and since x is approaching 4 from the left (less than 4, getting greater) both the numerator and denominator are going to be negative...do u get that?

    • 2 years ago
  68. abannavong Group Title
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    can u explain it a bit more im still a bit confused

    • 2 years ago
  69. jldstuff393 Group Title
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    sure. so the way i think of it, is to basically imagine x as some number a little less than 4.. 3.something. when you plug that number into the denominator, 3.something - 4 is always going to be negative. (anything less than 4) - 4 will be negative, right? (3-4=-1, 3.5-4=-0.5, 3.99-4 = -0.01)

    • 2 years ago
  70. abannavong Group Title
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    oh yeah so its like it keeps on approaching 4 right?

    • 2 years ago
  71. jldstuff393 Group Title
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    and for the numerator, you can imagine that since the square root of 4 is 2, the square root of anything less than 4 will be less than 2... when you subtract 2 from that number, the result is negative

    • 2 years ago
  72. jldstuff393 Group Title
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    ya it keeps approaching 4

    • 2 years ago
  73. jldstuff393 Group Title
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    but from the less-than-4 side because it has the minus sign (x->4-)

    • 2 years ago
  74. abannavong Group Title
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    basically the limit for it will negative right?

    • 2 years ago
  75. jldstuff393 Group Title
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    well the numerator and denominator are both negative, right?

    • 2 years ago
  76. jldstuff393 Group Title
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    so the fraction comes out positive

    • 2 years ago
  77. abannavong Group Title
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    oh ok

    • 2 years ago
  78. abannavong Group Title
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    why r limits soo confusing :3

    • 2 years ago
  79. jldstuff393 Group Title
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    lol i know. wanna see my limit problem that no one will answer?

    • 2 years ago
  80. jldstuff393 Group Title
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    Limit as x->(pi/2) of [tanx+1/(x-pi/2)]?

    • 2 years ago
  81. jldstuff393 Group Title
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    thats my question that no one will answer :(

    • 2 years ago
  82. jldstuff393 Group Title
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    i had it on a quiz today

    • 2 years ago
  83. abannavong Group Title
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    can u do the substitution method for tht one

    • 2 years ago
  84. abannavong Group Title
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    i think

    • 2 years ago
  85. abannavong Group Title
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    idk

    • 2 years ago
  86. jldstuff393 Group Title
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    no because tan of pi/2 is undefined, it goes to neg infinity and infinity from the right and left....

    • 2 years ago
  87. abannavong Group Title
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    oh yeah ur right oops lol

    • 2 years ago
  88. jldstuff393 Group Title
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    ya...it's brutal.

    • 2 years ago
  89. jldstuff393 Group Title
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    anyway

    • 2 years ago
  90. abannavong Group Title
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    but im still kinda confused on the continuity limits and one-sided ones

    • 2 years ago
  91. nj1202 Group Title
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    oh boy satellite's here.... it's about to get real xD

    • 2 years ago
  92. jldstuff393 Group Title
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    i just realized that your problem is tricky. you need to factor your x-4 into (sqrt(x)-2)(sqrt(x)+2) then cancel out the numerator

    • 2 years ago
  93. satellite73 Group Title
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    \[\lim_{ x\to\frac{\pi}{2}}\frac{\tan(x)+1}{x-\frac{\pi}{2}}\]?

    • 2 years ago
  94. jldstuff393 Group Title
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    oooh well that's my problem but not hers

    • 2 years ago
  95. jldstuff393 Group Title
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    mines posted

    • 2 years ago
  96. satellite73 Group Title
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    oh ok

    • 2 years ago
  97. jldstuff393 Group Title
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    somewhere down the list!

    • 2 years ago
  98. abannavong Group Title
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    @jldstuff393 how do u do my problem kinda confuzzled

    • 2 years ago
  99. jldstuff393 Group Title
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    i know i know. u remember difference of squares? like (x^2 - 4) = (x-2)(x+2)?

    • 2 years ago
  100. satellite73 Group Title
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    is this the problem \[\lim_{x \rightarrow 4-} \frac{ \sqrt{x}-2 }{ x-4 }\]?

    • 2 years ago
  101. abannavong Group Title
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    oh yeah i remember different of squares

    • 2 years ago
  102. abannavong Group Title
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    yeah @satellite73 thts my problem

    • 2 years ago
  103. jldstuff393 Group Title
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    so unless satellite has a less tricky way to solve it, you can factor (x-4) as a difference of squares... (sqrt(x)-2)(sqrt(x)+2)

    • 2 years ago
  104. jldstuff393 Group Title
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    does that make sense?

    • 2 years ago
  105. satellite73 Group Title
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    the gimmick here is to "rationalize the numerator" by multiplying by the conjugate of \(\sqrt{x}-2\) which is \(\sqrt{x}+2\)

    • 2 years ago
  106. jldstuff393 Group Title
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    cuz when u have that, you can cancel out the (sqrt(x)-2) in the numerator and denominator

    • 2 years ago
  107. abannavong Group Title
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    so u multpily the top and bottom by \[\sqrt{x} +2 \]

    • 2 years ago
  108. satellite73 Group Title
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    this works because \((a+b)(a-b)=a^2-b^2\) and so \((\sqrt{x}-2)(\sqrt{x}+2)=x-4\)

    • 2 years ago
  109. satellite73 Group Title
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    that cancels with the \(x-4\) in the denominator, which is what you want, you want to cancel the zero

    • 2 years ago
  110. abannavong Group Title
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    and x-4 can cancel in denominator

    • 2 years ago
  111. abannavong Group Title
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    oh ok!

    • 2 years ago
  112. abannavong Group Title
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    so ur limit is one right?

    • 2 years ago
  113. satellite73 Group Title
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    leaves you with \[\lim_{x \rightarrow 4} \frac{ \sqrt{x}-2 }{ x-4 }=\lim_{x \rightarrow 4} \frac{1 }{ \sqrt{x}+2}\]\]

    • 2 years ago
  114. abannavong Group Title
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    oh ok i get

    • 2 years ago
  115. satellite73 Group Title
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    no limit is not one, now you have to replace \(x\) by 4

    • 2 years ago
  116. abannavong Group Title
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    once u replace x with 4 u get 1/ 4 right?

    • 2 years ago
  117. satellite73 Group Title
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    yup

    • 2 years ago
  118. satellite73 Group Title
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    whenever you get \(\frac{0}{0}\) you need some gimmick to factor and cancel to get rid of the zeros

    • 2 years ago
  119. jldstuff393 Group Title
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    satellite would you mind helping me after you're done here? O.o good luck guys!

    • 2 years ago
  120. abannavong Group Title
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    oh ok

    • 2 years ago
  121. satellite73 Group Title
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    sometime sit is easy sometimes it is a pain. but you get used to it

    • 2 years ago
  122. abannavong Group Title
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    @satellite73 i have some more questions!

    • 2 years ago
  123. satellite73 Group Title
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    @jldstuff393 sure if i can what is the question

    • 2 years ago
  124. satellite73 Group Title
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    @abannavong go ahead and post them, i will look

    • 2 years ago
  125. jldstuff393 Group Title
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    i'll repost it, thanks!

    • 2 years ago
  126. abannavong Group Title
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    kk! thanks @satellite73

    • 2 years ago
  127. abannavong Group Title
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    14, 16, and 18

    • 2 years ago
  128. satellite73 Group Title
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    i can't really read it, but for 16 replace \(x\) by 2 in both formulas if you get the same number, that is the limit if you get different numbers, there is no limit

    • 2 years ago
  129. abannavong Group Title
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    number 14 its as deltax goes to 0 with the little plus sign

    • 2 years ago
  130. satellite73 Group Title
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    i can't read 18 either is it \(\lim_{ x\to 1^+}\) or \(\lim_{x\to 1^-}\)?

    • 2 years ago
  131. abannavong Group Title
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    x-> 1+

    • 2 years ago
  132. satellite73 Group Title
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    then replace \(x\) by 1 in the second formula you get \(1-1=0\)

    • 2 years ago
  133. abannavong Group Title
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    oh ok

    • 2 years ago
  134. abannavong Group Title
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    can u help with 14 also

    • 2 years ago
  135. abannavong Group Title
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    its confusing

    • 2 years ago
  136. satellite73 Group Title
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    sure, but i can't really read it can you post in a new question? you do not need to write \(\Delta x\) you can just write \(h\) instead

    • 2 years ago
  137. satellite73 Group Title
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    it is actually very straight forward we will do a little algebra and that is all

    • 2 years ago
  138. abannavong Group Title
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    kk hang on dinner quickly

    • 2 years ago
  139. satellite73 Group Title
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    \[(x+h)^2+x+h-(x^2+x)=x^2+2xh+h^2+x+h-x^2-x\] \[\cancel{x^2}+2xh+h^2+\cancel{x}+h-\cancel{x^2}-\cancel{x}=2xh+h+h^2\] divide by \(h\) and get \(2x+1+h\) let \(h\to 0\) be replacing \(h\) by 0 and you get \(2x+1\)

    • 2 years ago
  140. abannavong Group Title
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    oh ok! @satellite73 im back from dinner

    • 2 years ago
  141. satellite73 Group Title
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    that was a quick meal, you should take your time i wrote the solution above, but i used \(h\) instead of \(\Delta x\) it is mostly just algebra, to get rid of the factor of \(\Delta x\) in the denominator, then you replace it by 0

    • 2 years ago
  142. abannavong Group Title
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    i did eat dinner

    • 2 years ago
  143. abannavong Group Title
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    im eating and doing work lolz

    • 2 years ago
  144. satellite73 Group Title
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    you are either eating late or are on the left coast

    • 2 years ago
  145. abannavong Group Title
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    im on the west coast lolz

    • 2 years ago
  146. satellite73 Group Title
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    after you finish eating, read my solution above i hope all the steps are clear they are almost all algebra steps, only at the end do you replace \(h\) by 0

    • 2 years ago
  147. abannavong Group Title
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    kk

    • 2 years ago
  148. abannavong Group Title
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    oh ok its jst basic algebra right? so u foil the (x+h)^2 right?

    • 2 years ago
  149. satellite73 Group Title
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    yes exactly it is \((x+h)(x+h)=x^2+2xh+h^2\)

    • 2 years ago
  150. abannavong Group Title
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    oh ok and then from there u can combine the like terms and get 2xh+h+h^2

    • 2 years ago
  151. satellite73 Group Title
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    yes,

    • 2 years ago
  152. satellite73 Group Title
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    that is all in the numerator you get \[\frac{2xh+h+h^2}{h}\]

    • 2 years ago
  153. abannavong Group Title
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    and then can u factor out the h on the numerator

    • 2 years ago
  154. satellite73 Group Title
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    yes factor and cancel or divide each term by \(h\) it is the same thing

    • 2 years ago
  155. abannavong Group Title
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    oh ok and then u get 2x+1

    • 2 years ago
  156. satellite73 Group Title
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    yes

    • 2 years ago
  157. abannavong Group Title
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    oh ok! i get tht now

    • 2 years ago
  158. satellite73 Group Title
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    whew that was exhausting

    • 2 years ago
  159. abannavong Group Title
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    \[\lim_{x \rightarrow2 ^{+}}(2x-\left[ \left| x \right| \right]\] help im confused with this one also

    • 2 years ago
  160. satellite73 Group Title
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    is that an absolute value sign?

    • 2 years ago
  161. satellite73 Group Title
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    or is that the floor function?

    • 2 years ago
  162. abannavong Group Title
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    its like a bracket and then two straight lines inside

    • 2 years ago
  163. satellite73 Group Title
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    do you know what that function is called?

    • 2 years ago
  164. abannavong Group Title
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    ughhh i dont remember what it was called

    • 2 years ago
  165. satellite73 Group Title
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    it is either called "floor" or "greatest integer"

    • 2 years ago
  166. abannavong Group Title
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    is it kinda like a piecewise function or something like tht

    • 2 years ago
  167. abannavong Group Title
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    oh yeah i remember!

    • 2 years ago
  168. satellite73 Group Title
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    so if \(x>2\) but close to 2, then \([x]=2\)

    • 2 years ago
  169. satellite73 Group Title
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    so you have \[\lim_{x\to 2}2x-2=4-2=2\]

    • 2 years ago
  170. satellite73 Group Title
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    to be more precise \[\lim_{x\to 2^+}2x-2=4-2=2\]

    • 2 years ago
  171. abannavong Group Title
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    oh ok! at first it seemed confusing but oh ok now i get it!

    • 2 years ago
  172. satellite73 Group Title
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    whereas \[\lim_{x\to 2^-}2x-2=4-1=3\] because if \(x<2\) but close to 2, then \([x]=1\)

    • 2 years ago
  173. abannavong Group Title
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    oh ok! tht seems soo much easier now!

    • 2 years ago
  174. satellite73 Group Title
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    yes if you know what it means it is not hard here is a nice picture of \(y=2x-[x]\) http://www.wolframalpha.com/input/?i=2x-floor%28x%29

    • 2 years ago
  175. abannavong Group Title
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    oh ok! i get it! i have another question again hang on

    • 2 years ago
  176. abannavong Group Title
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    so it says find the x-values(if any) at which f is not continuous. which of the discontinuities are removable. f(x)= 1/ x^2+1

    • 2 years ago
  177. satellite73 Group Title
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    \[f(x)=\frac{1}{x^2+1}\] has no discontinuities, since it is a rational function and the denominator is never 0 since \(x^2\geq 0\) for any \(x\) you can see that \(x^2+1\geq 1\) no matter what \(x\) is, so this function has no discontinuities. as the denominator is never zero

    • 2 years ago
  178. abannavong Group Title
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    oh ok thank you @satellite73

    • 2 years ago
  179. satellite73 Group Title
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    you are welcome enough math, study something else or do something more constructive

    • 2 years ago
  180. abannavong Group Title
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    pretty much tht was my last problem! time to find a prayer for theology class! thank you @satellite73 and study for english test tomorrow on grapes of wrath

    • 2 years ago
  181. satellite73 Group Title
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    good luck with the joads

    • 2 years ago
  182. abannavong Group Title
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    lolz thanks i have to find the intercalary chapters to write about :3 i dont even know what those r

    • 2 years ago
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