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Homework Help: L'H again just checking

  1. Jun 21, 2005 #1
    Sorry I had to start a new thread here because the latex was just not generating on the other thread, I don't know if it is my browser or something else but I am having a very difficult day with the tex.


    \lim_{x \rightarrow 0} \frac {\sqrt(5x+9)-3}{x} [/tex]

    This gives 0/0 off the bat so apply L'H once I get:

    \lim_{x \rightarrow 0} \frac {(5x+9).5}{2} [/tex]

    Which plugs in nicely to give [itex] \frac {45}{2} [/itex]

    Did I do that all right? :blushing:

    Hi again can any latex specialists help me. My code definitely says the equation I wrote above is (sqrt(5x+9))-3/x but on my browser there is a red message saying the latex is not valid when I look at the thread and on my preview it is showing me a completely different other tex thing that I wrote two hours ago . :cry: I can't seem to avoid this problem, does anyone else see the mix up that I am seeing or is it just my own browser having a hissy fit?
    Last edited: Jun 21, 2005
  2. jcsd
  3. Jun 21, 2005 #2


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    There's a thread guiding how to use Latex on the General Physics board. Get a look at it.
    Are you asking for:
    [tex]\lim_{x \rightarrow 0} \frac{\sqrt{5x + 9} - 3}{x}[/tex]?
    If that's what you are asking for, then your answer is wrong. You can recheck your differentiation.
    Viet Dao,
  4. Jun 21, 2005 #3
    I am using it right Vietdao, here is my code I have removed the tags so that you can see it.

    \lim_{substack{x\rightarrow 0\} \frac {\sqrt{(5x+9)-3}{x}

    \lim_{\substack{x\rightarrow 0\} \frac {(5x+9).5}{2}

    \frac {45}{2}
  5. Jun 21, 2005 #4


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    You can click on all Latex images to see its code. Get a look at mine. Obviously, you don't need that '\substack'.
    Viet Dao,
  6. Jun 21, 2005 #5
    hmm yes I see that but it doesn't explain why my preview kept showing the log function I latexed hours ago fixing the errors was impossible with that going on.

    Well I do want to ask you another question but I am not sure if I want another snippy intolerant answer.

    Now I know I am an absent minded dumblebum sometimes, but I really love my calculus and I ask this hoping that people will appreciate that some of us did not have the luxury of the complete set of tutorials prior to this point.

    Thank you Viet Dao for pointing out that I missed something in my differentiation now that I have found it I need to ask for a little more help.
    I keep forgetting what x^(-3/2) translates into, I recall vaguely that it's something like 1/cubrt(x).

    Can someone explain to me some way to remember how these (-n/n) indexes translate. Memorising them is getting confusing.

  7. Jun 21, 2005 #6


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    [tex]a ^ {\frac{m}{n}} = \sqrt[n]{a ^ m}[/tex]
    [tex]a ^ {-m} = \frac{1}{a ^ m}[/tex]
    Hope you get it. :smile:
    Viet Dao,
    Last edited: Jun 21, 2005
  8. Jun 21, 2005 #7


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    My posts are snippy? I just want to guide you instead of giving you the answer... :cry:
    Viet Dao,
    Last edited: Jun 21, 2005
  9. Jun 21, 2005 #8
    Thanks :smile:

    Is that then:

    [tex]\lim_{x \rightarrow 0} \frac{(5x + 9)^{-3/2}*5} {2} = \lim_{x \rightarrow 0} \frac{5} {2\sqrt(5x + 9)^3}[/tex]


    Sorry if I upset you, really. Darn it its doing the preview thing again, I can't see what I have done wrong with the preview not working. Fixed it.

    The working above gives 5/54 is that what you got?
    Last edited: Jun 21, 2005
  10. Jun 21, 2005 #9


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    Hmm, it's incorrect.
    It should be:
    [tex]\frac{5}{2\sqrt{5x + 9}}[/tex]
    [tex](x ^ \alpha)' = \alpha x ^{\alpha - 1}[/tex]
    Viet Dao,
  11. Jun 21, 2005 #10
    :rofl: Ok maybe I need a headache tablet and a good nights rest.

    grr exams making me nutty.

    Ok thanks very much Viet Dao for being patient. I dont know why I got -3/2 for a-1 except maybe that I have worn down my brain a bit.

    the answer then is 5/6 yes?
  12. Jun 21, 2005 #11


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    Yes. Anyway, I'll tell you something. To check the lim x -> something answer. Plug the x = some number near something in your function, and if the result is near your answer. Then the chance that you are correct is quite high.
    Take this problem an example. Plug x = 0.001 in your calculator. The result will be very near "5 / 6".
    Viet Dao,
  13. Jun 21, 2005 #12


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    Here's an elegant non calculus way to do it

    [tex] \lim_{x\rightarrow 0}\frac{\sqrt{5x+9} -3}{x}\frac{\sqrt{5x+9}+3}{\sqrt{5x+9}+3} =\lim_{x\rightarrow 0}\frac{5x}{x\left(\sqrt{5x+9}+3\right)}=\frac{5}{6} [/tex]

  14. Jun 21, 2005 #13
    Trig limit

    Yeah, thankyou Daniel I deliberately didn't use the conjugates in order to practice L'Hopitals because I had some trouble with it earlier in the day.

    Could someone help me with this Trigonometric limit

    [tex] \lim_{t\rightarrow 0}\frac {1-cost^2} {t^2}[/tex]

    I realise that I want to manipulate it to look like

    [tex] \lim_{t\rightarrow 0} \frac {1-cost} {t} = 0 [/tex]

    somewhere, but with trig being a weak point for me I'm a bit put off by the squared variable, I have some idea what to do with trig functions all squared and what to do when they are multiplied by constants but I am a bit lost on the t^2. It might be ridiculously simple, but I need a pointer in the right direction. Thanks
    Last edited: Jun 21, 2005
  15. Jun 21, 2005 #14
    [tex] \lim_{x\rightarrow 0}\frac {1-cost^2} {t^2}[/tex]

    If thats [itex] cos^2t[/itex] as I think it is, then the numerator is a simple trig identity. From there its cakework.

    edit: Also you might mean the limit as t -> 0 :approve:
    Last edited: Jun 21, 2005
  16. Jun 21, 2005 #15

    Oh gosh I am so embarassed I don't know it and can't seem to find it, is it really that basic? :redface:
  17. Jun 21, 2005 #16
    No its cos t^2 not cos^2 t.

    Oh yeah Oops I cut and pasted

  18. Jun 21, 2005 #17
    Ugh, this is the reason we use parentheses, is it cos(t^2) or cos(t) ^2, as in cos(t) * cos(t)?
  19. Jun 21, 2005 #18
    Oh dear another ooops the actual question I am supposed to answer is

    [tex] \lim_{t\rightarrow 0}\frac {1-cost^2} {t^4}[/tex]

    note that the denominator is raised to power 4, since I don't really know what I should be using to solve this I don't know if the t^4 denominator makes any difference, but it probably does .

    Still bumbling

  20. Jun 21, 2005 #19

    cos (t^2) The way the difference is shown in my course is by saying

    cos^2 t for (cos t)^2
    cost^2 for cos (t^2)

    Sorry to confuse you

  21. Jun 21, 2005 #20
    In that case at a glance I would think the squeeze theorem would suffice.

    Also the substitution u = t^2 might make things a bit simpler.

    Another idea is multiply by the conjugate.
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