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Trig limit of type 0/0

  1. Nov 4, 2009 #1
    Here's the deal. The problem is this:

    [tex]\lim_{x\to1}{\frac{1+\cos{\pi x}}{\tan^2{\pi x}}}[/tex]

    No matter how I twist and turn the variables around, I can't get it to behave nicely. I end up with either (infty - infty) or (infty * 0) or similar.
    I can't use l'Hopital because it doesn't apply in this case.
    I've tried my luck with trig identities, but to no avail.
    Squeeze theorem doesn't really work either (that's my guess from looking at the graph), besides, I wouldn't know which functions to pick.

    From graphing it I know the answer to be 1/2, but I have no idea how to get there.
    All I need is a hint, guys. Please point me in the right direction!

    Thanks,
    -q

    EDIT: Wait, just had an idea. L'Hopital's rule didn't work because the deriv's were 0/0 as well. But technically I could recursively use it (i.e. 2nd deriv)...? Trying this now...
    EDIT2: Okay, that worked. Assuming I didn't make some stupid mistake... these trig deriv's tend to get quite long and messy, but I got the answer I was supposed to get. Is that just a fluke or did I solve it?
     
    Last edited: Nov 4, 2009
  2. jcsd
  3. Nov 4, 2009 #2
    you should be able to do this with just algebra, I quickly did it to see what you'd have to do. have you tried using the old multiply/divide trick?

    you should also avoid using L'hopital's rule unless they really want you to use it.. I think that ugly limit questions can be healthy practise for your algebra skills
     
  4. Nov 4, 2009 #3

    well, you could always show your steps.. try doing it without L'hopitals rule. if you're part of a class doing limits, sometimes they don't let you use it, I don't know if that's the case with you

    I can give you a start:

    we have 1+cos(pix) * cos(pix)/sin(pix), would 1-cos^2(pix) be useful?
     
  5. Nov 4, 2009 #4
    Thanks for your help, emyt. I double-checked the L'Hopital version and it worked, but your approach looks more elegant. I'll give it a shot tomorrow, but right now I need sleep... :)

    -q
     
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