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abannavong

  • 3 years ago

I need AP Calculus help!!

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  1. abannavong
    • 3 years ago
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    i will post the assignment soon!

  2. abannavong
    • 3 years ago
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    Help with all plz!

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  3. abannavong
    • 3 years ago
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  4. jldstuff393
    • 3 years ago
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    hi

  5. abannavong
    • 3 years ago
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    the circled questions btw

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

  7. jldstuff393
    • 3 years ago
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    i mean circled

  8. abannavong
    • 3 years ago
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    its numbers 10, 14, 16, 18, 22

  9. jldstuff393
    • 3 years ago
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    oh i was looking at the first one

  10. abannavong
    • 3 years ago
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    kk

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

  12. abannavong
    • 3 years ago
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    kk

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

  14. abannavong
    • 3 years ago
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    we dont have to do number 8 lol

  15. jldstuff393
    • 3 years ago
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    :\ it looked circled to me lol

  16. abannavong
    • 3 years ago
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    its only numbers 10, 14, 16, 18, 22

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

  18. abannavong
    • 3 years ago
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    oh yeah duh i forgot i do lolz

  19. jldstuff393
    • 3 years ago
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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?

  20. abannavong
    • 3 years ago
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    kk

  21. abannavong
    • 3 years ago
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    hang on

  22. abannavong
    • 3 years ago
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    number 8 its as x goes to infinity

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

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

  25. jldstuff393
    • 3 years ago
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    oh ok.

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

  27. abannavong
    • 3 years ago
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    i dont think so

  28. jldstuff393
    • 3 years ago
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    k.

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

  30. jldstuff393
    • 3 years ago
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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.

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

  32. jldstuff393
    • 3 years ago
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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

  33. jldstuff393
    • 3 years ago
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    so what would you have if you did that?

  34. abannavong
    • 3 years ago
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    hang on im kinda confused

  35. jldstuff393
    • 3 years ago
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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

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

  37. jldstuff393
    • 3 years ago
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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

  38. jldstuff393
    • 3 years ago
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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)

  39. abannavong
    • 3 years ago
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    oh ok

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

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

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

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

  44. jldstuff393
    • 3 years ago
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    yup

  45. abannavong
    • 3 years ago
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    oh ok now i get number 8

  46. jldstuff393
    • 3 years ago
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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.

  47. abannavong
    • 3 years ago
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    why can u?

  48. jldstuff393
    • 3 years ago
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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.

  49. abannavong
    • 3 years ago
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    im kinda confused on tht part

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

  51. abannavong
    • 3 years ago
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    oh ok

  52. jldstuff393
    • 3 years ago
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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.

  53. abannavong
    • 3 years ago
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    oh ok now i get it

  54. jldstuff393
    • 3 years ago
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    :) cool

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

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

  57. abannavong
    • 3 years ago
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    kk

  58. jldstuff393
    • 3 years ago
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    so i guess pick one for us to do

  59. abannavong
    • 3 years ago
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    kk

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

  61. jldstuff393
    • 3 years ago
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    k

  62. abannavong
    • 3 years ago
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    let me post the problem

  63. jldstuff393
    • 3 years ago
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    kk

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

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

  66. abannavong
    • 3 years ago
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    yeah

  67. jldstuff393
    • 3 years ago
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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?

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

  69. jldstuff393
    • 3 years ago
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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)

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

  71. jldstuff393
    • 3 years ago
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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

  72. jldstuff393
    • 3 years ago
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    ya it keeps approaching 4

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

  74. abannavong
    • 3 years ago
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    basically the limit for it will negative right?

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

  76. jldstuff393
    • 3 years ago
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    so the fraction comes out positive

  77. abannavong
    • 3 years ago
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    oh ok

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

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

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

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

  82. jldstuff393
    • 3 years ago
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    i had it on a quiz today

  83. abannavong
    • 3 years ago
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    can u do the substitution method for tht one

  84. abannavong
    • 3 years ago
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    i think

  85. abannavong
    • 3 years ago
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    idk

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

  87. abannavong
    • 3 years ago
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    oh yeah ur right oops lol

  88. jldstuff393
    • 3 years ago
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    ya...it's brutal.

  89. jldstuff393
    • 3 years ago
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    anyway

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

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

  92. jldstuff393
    • 3 years ago
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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

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

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

  95. jldstuff393
    • 3 years ago
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    mines posted

  96. anonymous
    • 3 years ago
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    oh ok

  97. jldstuff393
    • 3 years ago
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    somewhere down the list!

  98. abannavong
    • 3 years ago
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    @jldstuff393 how do u do my problem kinda confuzzled

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

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

  101. abannavong
    • 3 years ago
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    oh yeah i remember different of squares

  102. abannavong
    • 3 years ago
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    yeah @satellite73 thts my problem

  103. jldstuff393
    • 3 years ago
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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)

  104. jldstuff393
    • 3 years ago
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    does that make sense?

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

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

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

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

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

  110. abannavong
    • 3 years ago
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    and x-4 can cancel in denominator

  111. abannavong
    • 3 years ago
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    oh ok!

  112. abannavong
    • 3 years ago
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    so ur limit is one right?

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

  114. abannavong
    • 3 years ago
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    oh ok i get

  115. anonymous
    • 3 years ago
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    no limit is not one, now you have to replace \(x\) by 4

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

  117. anonymous
    • 3 years ago
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    yup

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

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

  120. abannavong
    • 3 years ago
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    oh ok

  121. anonymous
    • 3 years ago
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    sometime sit is easy sometimes it is a pain. but you get used to it

  122. abannavong
    • 3 years ago
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    @satellite73 i have some more questions!

  123. anonymous
    • 3 years ago
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    @jldstuff393 sure if i can what is the question

  124. anonymous
    • 3 years ago
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    @abannavong go ahead and post them, i will look

  125. jldstuff393
    • 3 years ago
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    i'll repost it, thanks!

  126. abannavong
    • 3 years ago
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    kk! thanks @satellite73

  127. abannavong
    • 3 years ago
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    14, 16, and 18

  128. anonymous
    • 3 years ago
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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

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

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

  131. abannavong
    • 3 years ago
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    x-> 1+

  132. anonymous
    • 3 years ago
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    then replace \(x\) by 1 in the second formula you get \(1-1=0\)

  133. abannavong
    • 3 years ago
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    oh ok

  134. abannavong
    • 3 years ago
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    can u help with 14 also

  135. abannavong
    • 3 years ago
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    its confusing

  136. anonymous
    • 3 years ago
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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

  137. anonymous
    • 3 years ago
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    it is actually very straight forward we will do a little algebra and that is all

  138. abannavong
    • 3 years ago
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    kk hang on dinner quickly

  139. anonymous
    • 3 years ago
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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 \(2x+1\)

  140. abannavong
    • 3 years ago
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    oh ok! @satellite73 im back from dinner

  141. anonymous
    • 3 years ago
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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

  142. abannavong
    • 3 years ago
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    i did eat dinner

  143. abannavong
    • 3 years ago
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    im eating and doing work lolz

  144. anonymous
    • 3 years ago
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    you are either eating late or are on the left coast

  145. abannavong
    • 3 years ago
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    im on the west coast lolz

  146. anonymous
    • 3 years ago
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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

  147. abannavong
    • 3 years ago
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    kk

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

  149. anonymous
    • 3 years ago
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    yes exactly it is \((x+h)(x+h)=x^2+2xh+h^2\)

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

  151. anonymous
    • 3 years ago
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    yes,

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

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

  154. anonymous
    • 3 years ago
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    yes factor and cancel or divide each term by \(h\) it is the same thing

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

  156. anonymous
    • 3 years ago
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    yes

  157. abannavong
    • 3 years ago
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    oh ok! i get tht now

  158. anonymous
    • 3 years ago
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    whew that was exhausting

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

  160. anonymous
    • 3 years ago
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    is that an absolute value sign?

  161. anonymous
    • 3 years ago
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    or is that the floor function?

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

  163. anonymous
    • 3 years ago
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    do you know what that function is called?

  164. abannavong
    • 3 years ago
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    ughhh i dont remember what it was called

  165. anonymous
    • 3 years ago
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    it is either called "floor" or "greatest integer"

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

  167. abannavong
    • 3 years ago
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    oh yeah i remember!

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

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

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

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

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

  173. abannavong
    • 3 years ago
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    oh ok! tht seems soo much easier now!

  174. anonymous
    • 3 years ago
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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

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

  176. abannavong
    • 3 years ago
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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

  177. anonymous
    • 3 years ago
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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

  178. abannavong
    • 3 years ago
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    oh ok thank you @satellite73

  179. anonymous
    • 3 years ago
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    you are welcome enough math, study something else or do something more constructive

  180. abannavong
    • 3 years ago
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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

  181. anonymous
    • 3 years ago
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    good luck with the joads

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

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