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Difficulty to find this limit

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello! I'm having difficulty to find this limit: [tex]\lim_{x\rightarrow -1}(\sqrt[3]{\frac{x^3+1}{x+1}})[/tex]


    2. Relevant equations



    3. The attempt at a solution
    This is what I'm trying to do to solve this limit: Let [tex]\frac{x^3+1}{x+1}=u[/tex] then
    [tex]\lim_{x\rightarrow -1}(\sqrt[3]{u}) = \lim_{x\rightarrow -1}(\sqrt[3]{-1})[/tex]
    I know something is wrong I'm just not sure about what is actually wrong. I'm thinking that [tex]\lim_{x\rightarrow -1}(\sqrt[3]{u})[/tex] Is wrong because that -1!

    Thanks!
     
    Last edited: Apr 22, 2012
  2. jcsd
  3. Apr 22, 2012 #2

    HallsofIvy

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    All you have done is "hide" the problem in that "u".

    I would think that your first step would be to write
    [tex]\frac{x^3+ 1}{x+ 1}= \frac{(x+1)(x^2- x+ 1)}{x+1}= x^2- x+ 1[/tex]
     
    Last edited: Apr 23, 2012
  4. Apr 22, 2012 #3

    SammyS

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    How did you come up with the above result?
    Factor x3 + 1 .
     
  5. Apr 22, 2012 #4
    Now I can see it!
    [tex]\lim_{x\rightarrow -1}f(x)=3[/tex] then [tex]\lim_{x\rightarrow 3}\sqrt[3]{u} =\sqrt[3]{3}[/tex] and that means [tex]\lim_{x\rightarrow -1}(\sqrt[3]{\frac{x^3+1}{x+1}})=\sqrt[3]{3}[/tex]
    Right?
     
  6. Apr 22, 2012 #5
    It's asking what the limit is when x approaches -1 not everything inside the radical. Simplify and cancel first then take the limit. Try factoring the denominator.

    edit: ^ you are making this way harder than it really is. Forget the limit, use algebra to simplify the expression then plug in -1 like you would in an function and see what you get.

    2nd edit: never mind, you got it
     
  7. Apr 22, 2012 #6

    Mark44

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    The last expression should be x2 - x + 1.
     
  8. Apr 23, 2012 #7

    HallsofIvy

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    Thanks! I'm too use to thinking of perfect squares!
     
  9. Apr 23, 2012 #8

    Mentallic

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    So you have the fraction [tex]\frac{x^3+1}{x+1}[/tex] and if you plug x=-1 into this, you get the indeterminate form 0/0. Remember that if you plug a value of x=a into a polynomial and it equates to zero, then it means that you have a root at x=a, and thus you can factor out (x-a) from the polynomial.

    This was the first step you needed to realize. Since plugging x=-1 into the numerator and denominator equates both of them to zero, then (x-(-1))=(x+1) is a factor of both the numerator and denominator (but in the denominator it's obvious) so you would then need to figure out how to factor x+1 out of the numerator.
     
  10. Apr 23, 2012 #9
    I'd like to say something real quickly. [itex]\displaystyle\lim_{x\to-1}\left(\sqrt[3]{u}\right)\neq\lim_{u\to-1}\left(\sqrt[3]{u}\right)[/itex]. By the same arguments, you could show that [itex]\displaystyle\lim_{x\to 0}\left(1\right)[/itex], and, by letting [itex]u=1[/itex], [itex]\displaystyle\lim_{x\to0}\left(u\right)=0[/itex], which is obviously not true.
     
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