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Find the limit of (x^7+1)/(x+1) as x goes to -1?

  1. Sep 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the limit of (x^7+1)/(x+1) as x goes to -1? Also find what this means for the derivative of that function. What is the slope at point "-1"?


    2. Relevant equations
    Use the difference of squares to factor and then cancel the bottom.



    3. The attempt at a solution
    I did this and then got x^6-x^5+x^4-x^3+x^2-x+1, but when I plug in -1, I get 7. Which has to be wrong because after graphing the function, the function's slope at -1 looks like it should be negative slope. Thanks for any help.
     
  2. jcsd
  3. Sep 6, 2010 #2
    Seems like you may have misinterpreted the limit here?

    You are correct, the limit as x goes to -1 is 7. This tells us that as we approach the value of x = -1 we find that your function is equal to 7. This has nothing to do with the value of the slope nor does it characterize the slope in any way.
     
  4. Sep 6, 2010 #3
    I guess I'm a little confused by the question the homework is asking. It says "After evaluating the limit. This derivative should tell the slope of the tangent line to the curve y = ? and x = ?"

    Should I be interpreting this limit as the "f ' (x) = f(x)-f(c)/x-c" then as x --> c should be the derivative. In that case, the function would x^7 and I'm evaluating the slope at -1. Which, after graphing it, looks more like 7 as the slope. Does this seem right?
     
  5. Sep 6, 2010 #4

    Dick

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    Yes, they gave you a difference quotient for f(x)=x^7 at x=(-1). And the slope there is 7.
     
  6. Sep 7, 2010 #5

    epenguin

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    Perhaps we should spell this more out because it is quite a tough problem, and my calcs. are not agreeing with yours.

    I am getting the slope at x = -1 to be -21. I get this both from an algebraic calculation, and from a graph of the derivative, and also reasonable near-agreement from taking a couple of points near x = -1 on the original function, graphed which is only a rough calc. but is not near -7.

    Work out the derivative of your (x7 + 1)/(x + 1). Graph the numerator and you will see what looks like a double root at x = -1. Result: you have a cancellation of the (x+1)2 factor of the denominator. The easiest way to see how this numerator factorises is by long division I find.
     
    Last edited: Sep 7, 2010
  7. Sep 7, 2010 #6

    Mark44

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    It seems to me that the wording in the OP might be different from the problem as given in the text, and as a result, there is ambiguity in what the function of concern actually is.

    As Dick points out, this limit --
    [tex]\lim_{x \to -1} \frac{x^7 + 1}{x + 1}[/tex]
    -- is the difference quotient for finding the derivative of f(x) = x7 + 1 at x = -1.

    Edit: I take that back. I should have said "is the difference quotient for finding the derivative of f(x) = x7 at x = -1."
     
    Last edited: Sep 7, 2010
  8. Sep 7, 2010 #7

    HallsofIvy

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    It is also, of course, the difference quotient for finding the derivative of [itex]f(x)= x^7[/itex] at x=-1 which is what Dick actually said. Since the two functions, [itex]x^7[/itex] and [itex]x^7- 1[/itex] differ by a constant, they have the same derivative at any x.
     
  9. Sep 7, 2010 #8

    Mark44

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    Right - I misspoke.
     
  10. Sep 8, 2010 #9

    epenguin

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    Your polynomial's value at -1 is 7. The slope there is -21.

    So a bit did I.
    I remember I was always struck by this fact that you could find the value, or strictly a limit value to f/g when both f and g are 0 for some value, a, of x, i.e. when f/g = 0/0, and it is f '(a)/g'(a). It seemed magic you could give definite significance to 0/0.

    But is was only something I ever saw out of the corner of my eye never looking carefully. Now it looks a bit trivial. I mean if that happens it just means f and g have a common factor (x-a). If f, g are polynomials you can always find and remove the common factor (in practice not just in theory). So no wonder the curve looks continuous with nothing special at the point. I wonder therefore what is the point of an exercise like this? Do you have for pure-mathematical doctrine respect the integrity of and g? Or is it a preparation for cases where f and g are not polynomials and you may not be able to eliminate the factor or not easily?

    Anyway, if instead of eliminating you do keep the factor, on differentiating you get a squared factor top and bottom as mentioned.

    You can solve your x^6-x^5+x^4-x^3+x^2-x+1 algebraically - can first reduce to a cubic in (x + 1/x) but I cannot see any special shortcuts after that. It looks to have no real roots. Its derivative looks to have only one real root about 0.678; I am tempted to think that too can be obtained algebraically but cannot see how.
     
  11. Sep 8, 2010 #10

    Hurkyl

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    The short-cut is to multiply by x+1.
     
  12. Sep 8, 2010 #11

    epenguin

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    :blushing: Stupid of me, it was staring me in the face. So the solutions are all the nonreal seventh roots of -1.

    Is solving its derivative almost equally easy? Or impossible?

    Edit: yes, should be able to do that. Roots of the polynomial so simply related should be able to do something with its derivative.
     
    Last edited: Sep 8, 2010
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