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Limit of ugly rational

  1. May 4, 2012 #1
    I have been an ugly rational function. I graphed it and put end behaviors and intercepts pods etc...

    Now after my teacher asked me to take the limit of this function and she has informed me that it is a much more simple way than taking the derivative of the entire peice.

    [(x+1)^2(x+3)^3(5-x)^7(x+9)(x-1)]\[(x-7)^3(x+4)(3-x)^2(25)


    So can some one give me a hint on how I can cancel pieces to simplify the expression ?

    I know numerator is degree 14 and denominator is degre but my teacher does not want me to do end behavior.

    Any help is appreciated
     
  2. jcsd
  3. May 4, 2012 #2
    You're trying to calculate the limit as x approaches what? That would help in knowing what you can cancel out.
     
    Last edited: May 4, 2012
  4. May 4, 2012 #3

    Mark44

    Staff: Mentor

    "take the limit of this function" - as x does what? When you post a problem, give us the full problem description.

    Also, do not just blow away the problem template, which includes the problem description, relevant formulas/equations, and your efforts.
     
  5. May 7, 2012 #4
    There is no problem description to begin with. it was an equation i was asked to graph it. I did, and thats not what i need help with...

    I'm sorry i didn't specify which part i needed to take the limit of, but i thought end behavior would imply that i was taking a limit as x approaches infinity from the positive and negative sides.


    So now that you've scolded me, could you help me out?

    How can i take this rational functions limit, and simplify it manually to a point in which i can prove the end behavior of this graph.
     
  6. May 7, 2012 #5

    micromass

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    The limit to infinity of a rational function is determined by the highest terms. So, for example:

    [tex]\lim_{x\rightarrow +\infty}{\frac{x^{100}+4x^5+6x}{5x^{100}+x^{50}}}[/tex]

    we can eliminate all terms but the highest one, so we get

    [tex]\lim_{x\rightarrow +\infty}{\frac{x^{100}}{5x^{100}}}=\frac{1}{5}[/tex]

    Try to do the same thing with your fraction.
     
  7. May 7, 2012 #6
    (5-x)^7 / (x-7)^3

    So I am allowed to do this with full peice brackets? if so please tell me what allows me to do this.. Like why?

    And if I do this I am still left with ugly degree of 7 which I'm not expanding.z
    I have a feeling, that the
    method you're talking about is only for standard form. (Which I can easily easily solve for.

    I am having a problem taking the limit of this equation while it is all in brackets due to the largeness and complexity.
     
  8. May 7, 2012 #7

    Mark44

    Staff: Mentor

    $$\frac{(5 - x)^7}{(x - 7)^3} = -\frac{(x - 5)^7}{(x - 7)^3}$$
    $$=-\frac{x^7(1 - 5/x)^7}{x^3(1 - 7/x)^3} = -\frac{x^7}{x^3}\cdot\frac{(1 - 5/x)^7}{(1 - 7/x)^3}$$

    Can you take it from there?
     
  9. May 7, 2012 #8
    Strictly speaking the way to go about this, using the method given by micromass, is to figure out what the highest exponent term would be if you had expanded it out fully, and neglect all the other terms. You can determine that by simply multiplying all the powers of [itex]x[/itex] in each term together (for example, the highest order term in the numerator would be [itex]-x^{14}[/itex]; you should be able to see this without expanding everything out).
     
  10. May 7, 2012 #9

    micromass

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    No, this is not correct. The method I suggested only works if the numerator and denominator is fully expanded. Here it is completely factored and the method does not work.
    So, you first need to expand the numerator and denominator completely. Of course, this can be done easily: you don't need to completely expand everything, you just need to find out the coefficient of the highest term. So if you're a bit smart, it's not much work.
     
  11. May 8, 2012 #10

    Ah this is marvelous, thank you very much sir.

    And yea i can take it from here.

    One last question, i wouldn't just take the highest degree peice and do this correct? i would do this for every single peice in the numerator and denominator of the equation, then combine the x values in the numerator + denominator?
     
  12. May 8, 2012 #11

    Mark44

    Staff: Mentor

    Yes, that is correct.
     
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