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Calculus 1 (limits, VA and constants)

  1. Jan 28, 2006 #1
    Im in calculus 1 and need some help

    Question: Find the vertical asymptotes for…
    F(x) = x^3 + 8 / x^2 –4
    I got a va of –2

    F(x) = 2x/ sin2x
    I got a va of 1

    Find the limit
    Lim tan^2 2x/ x^2
    x-0

    I got 4 as my answer

    Lim x^2 –4 / (x + 2 )^2
    x- (-2)

    I got –4/0 = no limit

    Constant problem

    F(x) = -x + 2 x<c
    2x^2 –2x –4 x_>c

    Not sure how to answer this.

    Where can I find notes on this section online?
    Thanks
     
  2. jcsd
  3. Jan 28, 2006 #2

    HallsofIvy

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    Did you notice this NOT the "homework" section?
    PLEASE use parentheses! I assume you really meant
    F(x)= (x^3+ 8)/(x^2- 4)= ((x+ 2)(x^2- 2x+ 4))/((x+2)(x-2))=
    ((x-2)(x-2))/(x-2)= x-2 as long as x is not 2 or -2. I don't see any vertical asymptote there.

    F(1)= 2/sin 2. How is that "vertical"?

    HOW did you get 4?

    Once again, HOW did you get "-4/0"?

     
  4. Jan 28, 2006 #3

    VietDao29

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    Homework Helper

    As a matter of fact, [itex](x ^ 2 - 2x + 4) \neq (x - 2) ^ 2[/itex] :wink:
    I think he probably meant:
    [tex]\lim_{x \rightarrow -2} \frac{x ^ 2 - 4}{(x + 2) ^ 2} = \lim_{x \rightarrow -2} \frac{x - 2}{(x + 2)} \rightarrow \frac{-4}{0}[/tex]
    (@ gator, I wouldn't say -4 / 0 = no limit (as -4 / 0 is not really well-defined), I'd rather say, as x tends to -2, the numerator tends to -4, and the denominator tends to 0, hence there's no limit).
    How much will a chance probably cost me? :tongue:
     
    Last edited: Jan 28, 2006
  5. Jan 29, 2006 #4

    HallsofIvy

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    Well, dang me! It was probably that fourth whiskey and soda.
     
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