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Math Help - Limits and inverse trig functions

  1. #1
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    Limits and inverse trig functions

    Hi everyone,


    I'm currently studying for my calculus test tommorow and I have some problems which I have no clue how to solve. One problem is a limit and the other one is about inverse trig functions.


    1) Lim (cos 2x)^(csc^2 2x)


    x-> 0


    2) 2. Cos(2arccotx)


    Any help at all is very much appreciated, as I truly have no idea how to get started on especially the first exercise. Oh and I should mention that we are not allowed to use L'Hopitals rule for these problems.
    Last edited by tonykart44; November 8th 2012 at 11:25 AM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Limits and inverse trig functions

    1) \lim_{x \to 0} \frac{\cos 2x}{\csc^2(2x)}
    Note that the limit is equal to
    \lim_{x \to 0} \cos(2x) \sin^2 (2x)
    2) 2 \cos[2\mbox{arc}\cot(x)]
    Hint: use the double-angle identity for \cos(2u)
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    Re: Limits and inverse trig functions

    Thank you very much for your fast answer! But I think I wasn't clear on the first problem, It should be cos(2x) to the power csc^2 2X Sorry if that wasnt clear.

    For the second one I did what you said and I got (2X^2/(1+X^2)) -1 does that look alright to you?
    Last edited by tonykart44; November 8th 2012 at 11:56 AM.
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Limits and inverse trig functions

    No, it's my fault, for some reason I red a fraction in stead of a power.
    In case we have to compute \lim_{x \to 0} \cos(2x)^{\csc^2(2x)}. Use the fact that e^{\ln(x)}=x thus we can write the limit as \lim_{x \to 0} e^{\csc^2(x) \ln[\cos(2x)]} (note \ln(a^b) = b \ln(a)). Because the exponential function is continuous we can write the limit as e^{\displaystyle \lim_{x \to 0} \frac{\ln[\cos(2x)]}{\sin^2(2x)} which can be solved by using l'Hopital's rule.

    The solution of the second question is correct. (I assume the 2. was not a factor 2 but just an indication of question 2)
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  5. #5
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    Re: Limits and inverse trig functions

    Once again, I will need to annoy you because for these problems, we are actually not allowed to use L'H˘pitals rule. I'm trying to use trig functions to get around that, but I just can't seem to get that right.

    Ok so I got this one in the form (1-2sin^2 x)^(1/2sin^2 2x) but i don't know how to go on from there. I know i'm almost there but i don't know how to get rid of the 2x above, I know it's silly, but i just don't see it.
    Last edited by tonykart44; November 8th 2012 at 12:34 PM.
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Limits and inverse trig functions

    I have no idea how to solve this without l'Hopital's rule.
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