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Finding limits without use of l'Hôpital's rule.

  1. Oct 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi.

    I have a problem finding the limit of two different problems - without use of l'Hôpital's rule. I only know how to do this with use of the l'Hôpital's rule, therefore I'm seeking help to solve this problem.


    2. Relevant equations

    The problems are:

    Determine the limits without using l'Hopital's rule:

    lim (x-2)/((√(x^2-3)-x/2)
    x->2

    lim (2x-√(4x^2+3x))
    x->∞

    3. The attempt at a solution

    First thing first, the 1st problem.

    I do not have any attempt at a solution yet, as I do not know how to go about to solve this problem. I should add that I have solved this problem graphically, and I know that the answer should be 2/3.

    Should I attempt to simplify the expression (I'm a bit rusty on that, unfortunately), factorize the expression or how should I go about to solve this?

    I'm a bit clueless here, but I have my books open, ready to make an effort into completing this problem.

    Thanks for any input.
     
  2. jcsd
  3. Oct 12, 2013 #2
    I do know that I'll have to do something with the expression in order to be able to cancel something out with each other.

    My best attempt so far is to multiply both the numerator and denominator by √(x^2-3), but I'm not too sure on how I write that out.
     
  4. Oct 12, 2013 #3
    Can anyone at least point me towards how I can find the solution to this problem myself?
     
  5. Oct 12, 2013 #4

    vanhees71

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    I don't know, how to do the work without de L'Hospital's rule. The only idea I have is to use series expansions, but in fact that's also de L'Hosptital's rule, but it's hidden behind its proof ;-).

    E.g., for the first problem
    [tex]\sqrt{x^2-3}-\frac{x}{2}=\frac{3}{2}(x-2)+\mathcal{O}[(x-2)^2].[/tex]
     
  6. Oct 12, 2013 #5
    Thanks a lot for showing interest in my topic, even if I haven't got any solid work to analyze.

    I do, however, doubt that this is how I should solve this problem, because we haven't learned anything similar to this I'm afraid. To be honest, I don't know what technique I'm supposed to use for this problem.

    Would it be possible to factorize/simplify the expression, cancel out identical expressions, and then swap 2 for x, and get 2/3 as an answer? I've tried, but I can't figure out how to factorize/simplify it. I'm currently looking at how to simplify and factorize expressions like this, so I post any solutions or progress I find.
     
  7. Oct 12, 2013 #6

    vanhees71

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    Ok, for the first problem, you can also try to rationalize the denominator, using
    [tex](\sqrt{x^2-3}-x/2)(\sqrt{x^2-3}+x/2)=(x^2-3)-x^2/4=3/4 x^2-3.[/tex]
     
  8. Oct 12, 2013 #7
    For the first question change the denominator to

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

    Then multiply the top and bottom by

    [tex]\sqrt{x^2 - 3} + \frac{x}{2} [/tex]

    Then by doing some factorization of the top and bottom and some cancellations of terms you should get the desired result. I scribbled it out an a sheet of paper here and it seems to works out.

    I haven't had a chance to try the second one yet... Maybe when I get back I'll give it a shot. In any event for the first no series expansions or approximations, etc. are required. Just some good ol; fashioned elementary algebra.

    Hope this helps.
     
  9. Oct 12, 2013 #8

    Mark44

    Staff: Mentor

    For the second problem, multiply numerator and denominator by 1 in the form of 2x + √(4 x2 + 3x) over itself. After doing some careful factoring of the denominator, you should be able to evaluate your limit.

    BTW, it's against Physics Forums rules to "bump" your thread sooner than 24 hours after you post it.
     
  10. Oct 12, 2013 #9
    Hint... for the second limit rationalize by multiplying by

    [tex] \frac{2x + \sqrt{4x^2 + 3x}}{2x + \sqrt{4x^2 + 3x}}[/tex]

    So...

    [tex](2x - \sqrt{4x^2+3x})\frac{2x + \sqrt{4x^2 + 3x}}{2x + \sqrt{4x^2 + 3x}} [/tex]

    Then proceed from there with. Factor when possible and simplify radical expressions. Give it a try and let us know how it works out.
     
    Last edited: Oct 12, 2013
  11. Oct 12, 2013 #10
    Thanks a lot for this! I've been away for awhile, but I'm back now for a couple of hours of math problem solving.

    This is what I've gotten so far.

    [itex]\frac{(x-2)(\sqrt{x^2-3}+\frac{x}{2})}{(\sqrt{x^2-3}-\frac{x}{2})(\sqrt{x^2-3}+\frac{x}{2})}[/itex]

    = [itex]\frac{(x-2)(\sqrt{x^2-3}+\frac{x}{2})}{(\sqrt{(x^2-3) ^2}-\frac{x}{2}^2}[/itex] <--- the (x/2)^2 should be like this. Not (x^2/2).

    = [itex]\frac{(x-2)(\sqrt{x^2-3)}+\frac{x}{2})}{(x^2-3)-(\frac{x^2}{4})}[/itex]

    Where am I going wrong?

    Thanks a lot for all the help, and I apologize for not following the rules and spamming the thread earlier. I will look at the other inputs when I'm done with this first problem. I can't handle several problems at once I'm afraid.
     
  12. Oct 12, 2013 #11

    Mark44

    Staff: Mentor

    Your work looks fine so far. To continue, combine the two x2 terms in the denominator and factor the resulting binomial.
     
  13. Oct 13, 2013 #12
    I thought I was heading for trouble, thank you!

    [itex]\frac{(x-2)(\sqrt{x^2-3}+\frac{x}{2})}{\frac{3x^2}{4}-3}[/itex]

    [itex]\frac{4(x-2)(\sqrt{x^2-3}+\frac{x}{2})}{3(x+2)(x-2)}[/itex] (3x^2-12 rewritten)

    [itex]\frac{4(\sqrt{x^2-3}+\frac{x}{2})}{3(x+2)}[/itex]

    [itex]\frac{2\sqrt{x^2-3}+x}{3(x+2)}[/itex]

    When I now insert 2 for x, I get [itex]\frac{2}{3}[/itex], thank you for all your help :) If this looks ok now, I will be moving on to the next problem.

    By the way, it's a paper I'm doing atm. Can I post my next problem in the same post, or do I have to make a new post for another problem?
     
  14. Oct 13, 2013 #13

    arildno

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    Make a new thread.
     
  15. Oct 13, 2013 #14
    This is what I've got so far.


    (2x-[itex]\sqrt{4x^2+3x})[/itex][itex]\frac{2x+\sqrt{4x^2+3x}}{2x+\sqrt{4x^2+3x}}[/itex]

    = [itex]\frac{4x^2-4x^2+3x}{2x+\sqrt{4x^2+3x}}[/itex]

    =[itex]\frac{3x}{2x+\sqrt{4x^2+3x}}[/itex]

    Have I gone by it the wrong way? I feel like in order to continue this problem, I need to do the same thing in the denominator now - multiply it with itself or itself with - instead or +.

    Thanks for any input.
     
  16. Oct 13, 2013 #15

    Dick

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    ##\sqrt{4x^2+3x}=\sqrt{x^2 (4+3/x)}=\sqrt{x^2} \sqrt{4+3/x}##. See where to go from here?
     
  17. Oct 13, 2013 #16
    Unfortunately, no. I understand that you're rewriting the root in the denominator, but when I try to continue what you've done, I just end up with wanting to multiply by x, and that's just a step backwards, isn't it?

    Thanks a lot for your reply, I really appreciate the help.
     
  18. Oct 13, 2013 #17

    arildno

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    How much does 3/x measure relative to 4 when x is really big?
     
  19. Oct 13, 2013 #18
    0, I think, but I can't see how that's helping me right now, sorry!

    Hmm, can I do it like this?

    [itex]\frac{3x}{\sqrt{x^2}\sqrt{4+\frac{3}{x}}}[/itex]

    [itex]\frac{3x}{x\sqrt{4+\frac{3}{x}}}[/itex]

    Well, and then get something, something, [itex]\frac{3}{4}[/itex]

    I'm not sure what I was thinking...
     
  20. Oct 13, 2013 #19

    arildno

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    Well, 3/x becomes totally irrelevant towards 4, so that the MAIN contribution from your second term will be [itex]\sqrt{x^{2}}\sqrt{4}=2x[/itex]
    Since this main contribution does NOT cancel with the first term in the denominator (ALSO 2x), the total main component of the denominator=2x+2x=4x (in addition, you have some tiny nonsense that goes to zero as x goes to infinity)
     
  21. Oct 13, 2013 #20
    By following this logic, I get that the limit as x approaches infinity is 3/2. My graph tells me it should be -3/4.
     
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