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Limit homework help question

  1. Sep 19, 2011 #1
    1. The problem statement, all variables and given/known data

    712p91.jpg

    2. Relevant equations



    3. The attempt at a solution

    I need to make sure what Im doing is correct. My teacher hasnt been very helpful.
    In the first two parts iwe merely placed in the limit value and gotten ratios which I think are the limits.

    In part c and onwards iwe generally gotten an indeterminate form each time Iwe placed in the limit so i used L hopitals rule and differentiated the top and bottom functions and then placed in the limit value on i.e f'(a)/g'(a) and used that as my answer. Is my thought process correct?

    The answers I achieved are
    a 2
    b 272/11
    c 21.6
    d -1/6
    e 1/3
    f 1
    g 3

    Some of these are scary fractions and decimals. Should I be worried about those. comments and criticism is welcome. Thanks in advance!
     
  2. jcsd
  3. Sep 19, 2011 #2
    Re: Limits.

    c,e,f, and g are incorrect. g actually has 2 cases (or is undefined depending on how you look at it).

    If you show your work we can figure out what happened.
     
    Last edited: Sep 19, 2011
  4. Sep 19, 2011 #3
    Re: Limits.

    C. i differentiated the top and the bottom. So I got
    4x^3/(4x-5)
    Next I put in the value 3 in place of x. Okay wait so Im getting 108/7 is this correct?

    E.
    Again I differentiate.
    I get 1/(1/3x^-2/3) Put 27 in for x and I get 27. Is this the correct answer?

    F.
    Initially what I did was cancel out the x from all the terms.
    And Im left with

    (x^2 -7)/x^2 but i notice that thats not getting me any where

    so i differentiated top and bottom from the beginning.
    So I got (3x^2 -7)/3x^2

    this wasnt getting me anywhere so I differentiated this again
    and got 6x/6x. Here Im left with 1 hence my answer.

    G.
    I again did the same.
    (2x+1)/1 and put in x as 1. to get 3.

    Please let me know what i did wrong.

    Also a question regarding L hopitals rule its NOt only applicable when x tends to zero right? Because Iwe used it in cases where x has not tended to zero as well.
     
  5. Sep 19, 2011 #4

    symbolipoint

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    Re: Limits.

    Item #c can be simplied and then direct substitution of x=3 can be done. No need for L'Hopital's Rule.

    [itex]Lim_{x\rightarrow 3}\frac{x^4-81}{2x^2-5x-3}[/itex]

    Working with the expressin, [itex]\frac{(x^2+9)(x^2-9)}{(2x+1)(x-3)}[/itex]
    [itex]\frac{(x^+9)(x+3)(x-3)}{(2x+2)(x-3)}[/itex]
    [itex]\frac{(x^2+9)(x+3)}{(2x+1)}[/itex]
    Now you can substitute 3 for x in this last expression and evaluate.
     
  6. Sep 19, 2011 #5

    symbolipoint

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    Re: Limits.

    #f, L'Hopital's Rule gives these expressions,

    First derivative of top and bottom:
    [itex]\frac{3x^2-7}{3x^2}[/itex]
    Second derivatives...
    [itex]\frac{6x-0}{6x}[/itex], which is clearly equal to 1 (that is, as x approaches 0).
     
  7. Sep 19, 2011 #6
    Re: Limits.

    Thanks for the help. Can someone also look into e and g since Iwe been told that those are incorrect. My answers for e and g are 27 and 3 respectively. Thanks.
     
  8. Sep 19, 2011 #7

    eumyang

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    Re: Limits.

    Looks like e) is right. You can also try using the difference of cubes formula
    a3 - b3 = (a - b)(a2 + ab + b2).
    a, in this case, would be x1/3.
     
  9. Sep 19, 2011 #8
    Re: Limits.

    e looks good this time around.


    Try graphing out g and check out what the function does as x approaches 1. In a situation like this you can break the limit into two cases and specify what the limit is as x approaches 1 from both the positive and negative side. Some professors simply call it undefined and leave it at that though.
     
  10. Sep 19, 2011 #9
    Re: Limits.

    After you take the first derivatives of the num/den there you no longer have an indeterminate form and cannot apply L'hopital's Rule. Unless there is some other case which allows the second derivative to be taken?

    I don't believe that this limit is 1.
     
  11. Sep 19, 2011 #10
    Re: Limits.

    If you get an indeterminate form when you take the ration of the derivatives of the two function you can consider the second and third derivatives as well.
     
  12. Sep 19, 2011 #11

    eumyang

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    Re: Limits.

    Instead of taking the 2nd derivative, take the 1st derivative:
    [itex]\lim_{x \rightarrow 0} \frac{3x^2 - 7}{3x^2}[/itex]
    ... and split into a difference of two fractions. Remember that the limit of a difference is the difference of the limits. See what happens from there.
     
  13. Sep 19, 2011 #12

    symbolipoint

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    Re: Limits.

    My efforts on #f are way off. Not yet sure what the best method is, but graphing the function shows moving toward negative infinity as x approaches 0. No limit.
     
  14. Sep 19, 2011 #13
    Re: Limits.

    Yes but the problem in question did not result in an indeterminate form. The initial equation resulted in a type 0/0 ind. allowing L'hopital's rule to be applied. If you tried to take the limit again you would end up dividing by zero, but the numerator was a non-zero. I'm pretty sure that L'Hopital does not allow this by definition, as it is not an indeterminate form:

    http://en.wikipedia.org/wiki/L'Hôpital's_rule#Other_indeterminate_forms


    To solve limit f you might try taking the first derivative (L'hopital) as to end up with:

    [tex]lim_{x \to 0} \frac{3x^{2}-7}{3x^{2}}[/tex]

    From there, I broke the fraction into parts:

    [tex]1-\frac{7}{3x^{2}}[/tex]

    Now I simply factored out the 3 from the denominator and moved the limit on inside so that:

    [tex]1- \frac{7}{3(lim_{x \to 0}x^2)}[/tex]

    Then just use the fact that [itex](lim_{x \to 0}x^{n}) = (lim_{x \to 0}x)^{n}[/itex]

    Then you can just think about how when x goes to, but never "reaches" 0. the denominator get's really really small. Making the fraction tend to infinity. But it's one minus the fraction, so it shoots off to - infinity.
     
    Last edited: Sep 19, 2011
  15. Sep 19, 2011 #14

    uart

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    Re: Limits.

    Use L'Hopital's once and the numerator is -7 while the denominator is zero. So yes, it doesn't reach a limit.
     
  16. Sep 20, 2011 #15
    Re: Limits.

    wow thankyou. And thankyou everyone else too!
     
  17. Sep 20, 2011 #16

    verty

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    Re: Limits.

    L'hopital's rule has three conditions, and if they don't apply, you can't use it. You must always check them.

    (3x^2 - 7) / 3x^2 fails the first condition in the above question, the numerator and denominator do not have the same limit in that case.
     
  18. Sep 20, 2011 #17
    Re: Limits.

    what are those three conditions?
    As far as Im concerned all Iwe been doing is checking for an indeterminate form and if its tjere applying L hopitals rule.

    Also. When you guys suggested breaking it into two parts and not going for the second derivative, I remember my teacher saying somewhere that if the ratio of the first derivatives of the two function is zero you consider the ratio of the second derivatives? Is that incorrect?
     
  19. Sep 20, 2011 #18

    dynamicsolo

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    Re: Limits.

    This is not described as two "cases" generally. What you describe is that the two "one-sided limits" do not agree, so the (two-sided) limit does not exist. It is not simply "undefined" because a standard "epsilon-delta" argument would fail for this function near x = 1.

    Another way you can see that there's going to be trouble is that this function has an oblique asymptote of y = x + 1 and a vertical asymptote at x = 1 . For the graph to be contained by those lines, the curve for x > 1 must run off to "positive infinity" as x approaches 1 , while the branch for x < 1 must go to "negative infinity". So there is no meaning for a limit as x approaches 1.
     
    Last edited: Sep 20, 2011
  20. Sep 20, 2011 #19

    dynamicsolo

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    Re: Limits.

    If this is (f) you're talking about, you have to try something else because after taking first derivatives, the "limit" as x --> 0 is now "-7/0", so L'Hopital will avail you no further...
     
  21. Sep 20, 2011 #20
    Re: Limits.

    I haven't got into the epsilon-delta limit stuff. My professor used to have us break those limits into two cases, for when the variable is approaches fromt the positive and the negative. I'm sure you are probably right though.
     
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