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Question on Limit of a given function

  1. Jun 19, 2008 #1
    limx->pi/4 {tan(x)-1}/(pi-4x)


    I tried to solve it like this:
    lim x->pi/4 tan(pi-4x)=

    lim x->pi/4 tan(pi-5x+x)

    = limx->pi\4 {tanx+tan(pi-5x)}/{1-tan(pi-5x)tanx}
    Now
    lim x->pi/4 tan(pi-4x)=(pi-4x)
    Therefore,
    lim x->pi/4 (pi-4x)= lim x->pi/4 (tan(x)-1)/2 [Valid as lim fx/gx=limfx/limgx and lim (fx-gx)=limfx-limgx]

    hence
    lim x->pi/4 {tan(x)-1}/{pi-4x}=2
    But this is wrong
    the answer is 1/2


    Could someone please point out the mistake in my procedure
    Isnt the statement lim x->a (fx -gx)=lim x->a fx-1 where g(a)=1 coorect?

    Thanks
     
  2. jcsd
  3. Jun 19, 2008 #2
    Can you guys use Lhopital's rule yet? This limit is a good candidate for Lhopital.

    I tried following your steps but I don't get it.
     
  4. Jun 19, 2008 #3
    This question can be solved using L hopitals easily but i dont think that would help in developing mathematical intellect.But i do acknowledge that it is a beautiful rule.

    Which part do you not understand in my method?
     
  5. Jun 19, 2008 #4

    matt grime

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    There's no need to take a limit there. The answer is zero. That's the whole point of the question.



    Since you aren't actually taking any limits, and you're doing a trig identity thing why not drop the lims? It would make it easier to read.

    This doesn't make sense: the LHS is a limit (i.e. a real number if it exists). The RHS is an expression in x.


    Again, that limit is just 0.

    The first rule is *only* valid if the limit of g(x) at the point in question is *not* zero. The whole point of this question is that g(x) is zero at pi/4.



    Surely, from what you've written you've just claimed that the limit is 1? What you did is incorrect, though.

    If you want to do it by elementary means, then write tan as sin/cos, and attempt to find some non-calculus based proof about the limit of sin(y)/y as y tends to zero and adapt it.
     
  6. Jun 20, 2008 #5

    Gib Z

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    I'm confused. Isn't the question [tex] \lim_{x\to \pi/4 }\left( \frac{ \tan x -1}{\pi/4 - 4x} \right)[/tex], which in fact does need the limit?
     
  7. Jun 20, 2008 #6

    matt grime

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    The denominator is pi - 4x.
     
  8. Jun 20, 2008 #7

    Gib Z

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    O Sorry my bad. But it remains that it does need the limit. It's of the form 0/0.
     
  9. Jun 22, 2008 #8
    Mond, what you seem to be missing is that the limit of the numerator is zero and the limit of the denominator is zero. I think I understand your method. I think you incorrectly evaluate the limits getting a non-zero value for at least one of them.

    If you don't like L'Hopital's rule, you could do the same thing with a Taylor expansion.

    If you don't like Taylor expansions, you can do it with the binomial theorem. BTW, that's how Newton "got at" his concept of the calculus. He got at it through the means of the binomial theorem.

    You might have a little problem applying the binomial theorem to the tangent function. You might check out the general approach to working with trig identities found in most beginner expositions of elliptic integrals and the Weirstrauss parameterization of elliptic curves. One trick is to work with to the arctangent and then use the formular for the derivative of an inverse function (it is the reciprocal of the derivative of the function). Another is to use half angle forrumlae.

    Deacon John
     
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