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Finding the limit of a very complicated trig function + some extra questions

  1. Nov 29, 2011 #1
    1. The problem statement, all variables and given/known data

    I am trying to find the limit of the following question:

    lim (as x approaches infinite) sin[ arctan(((x^3)+2)/((x^2)+5)) + (((cosx)^2)/((|x|)^(1/2)))*arctan(2x)]

    (to see the above function better, without a million brackets, please use the following link)

    http://www.wolframalpha.com/input/?i=sin[+arctan%28%28%28x^3%29%2B2%29%2F%28%28x^2%29%2B5%29%29+%2B+%28%28%28cosx%29^2%29%2F%28%28|x|%29^%281%2F2%29%29%29*arctan%282x%29]


    2. Relevant equations



    3. The attempt at a solution

    My attempt so far is practically non-existent! This question is part of the harder set of exercises available from my university's department for first year calc, but I figure that if I can get to a point where I can solve these kinds of problems, I should have no problem on the test!

    OK, so before I even start describing my attempt, I have a quick question. Say we were given something like this:

    Find lim as x approaches infinity of sin(cos(f(x))), where f(x) is of the form 0/0 or infinity over infinity, is it correct to simply isolate f(x), use hoptals rule until you come to a conclusion, and then plug it back into the limit?

    Also, since the orignial question is the limit as x approaches POSITIVE infinity, can we just conclude that the |x|=x?

    OK so back to the attempt, if we truly are aloud to use Hopitals rule inside of trig functions, then I get the first arctan(f(x)) to be equal to pi/2. Is this correct so far?

    As for the second term within the sin, I cannot solve this! I tried making |x|=x and combined it into a single fraction. But I seem to get 0/infinity and I cannot find out how to simplify this so that it is of a form that I can use with Hopital's rule.

    I realize that this is a very loaded question, but thank you so much for your help!
     
  2. jcsd
  3. Nov 29, 2011 #2

    LCKurtz

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    You are on the right track by looking at the pieces from the inside out. Yes, the arctan terms go to π/2 because their arguments go to +∞. Looking at that last term, you have the arctan going to π/2, the cos2(x) term between 0 and 1, and a √(x) in the denominator. So what does that part approach? Put it all together and what does the sine approach? Does it agree with Wolfram's picture?

    Note that by taking the limits inside the functions, you are using the fact that all those functions are continuous.
     
  4. Nov 29, 2011 #3
    I strongly suggest you try and use the forums [tex] tags to typeset anything that looks that nasty, because it is really confusing to try and parse it. It'll be a worthwhile skill to have in the long run and you'll get more responses. LaTeX help

    What you need here is how to evaluate the limit of a composite function. The idea is that if the limit of f(g(x)) converges, then the limit of g(x) must converge. In particular, if (for simplicity suppose f and g are continuous) [itex]\lim_{y \to \alpha}f(y)=A[/itex] and [itex]\lim_{x \to \beta}g(x)=\alpha[/itex] then

    [tex]\lim_{x \to \beta}f(g(x)=\lim_{x \to \alpha}f(y)=A[/tex]

    So what you will do with this problem is evaluate the inner limits, which will, in short, allow you to say the lim Sin[~] = Sin[lim ~], where ~ is just the stuff inside. You'll be doing that a couple of times since your trig function is multiply composite. Also note that since x→∞, |x|=x.

    As an example, [itex]\lim_{x \to \infty}\cos(\frac{\sin(x)}{x})=\lim_{x \to 0}\cos(x)=1[/itex] since [itex]\lim_{x \to \infty}\frac{\sin(x)}{x}=0[/itex].
     
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