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Find the limit using taylor series

  1. Jan 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Using the taylor series at point ##(x=0)## also known as the meclaurin series find the limit of the expression:
    $$L=\lim_{x \rightarrow 0} \frac{1}{x}\left(\frac{1}{x}-\frac{cosx}{sinx}\right)$$
    2. Relevant equations
    3. The attempt at a solution

    ##L=\lim_{x \rightarrow 0} \frac{1}{x}\left(\frac{sinx-xcosx}{xsinx}\right)##
    ##L=\lim_{x \rightarrow 0} \frac{1}{x^2}\left(\frac{sinx-xcosx}{sinx}\right)##
    I need help to continue. This is the first problem im doing that asks me to claculate the limit using the taylor series. Am i suppose to write the taylor series for ##sinx,cosx## up to some element and somehow use them?
     
  2. jcsd
  3. Jan 2, 2017 #2

    Orodruin

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    Yes. Keep the leading non-zero term in ##x## in both denominator and numerator. The rest will be irrelevant.
     
  4. Jan 2, 2017 #3

    fresh_42

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    Shorter is directly the series for cotangent.
     
  5. Jan 2, 2017 #4

    vela

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    You may also find the series ##\frac{1}{1-z} = 1+z+z^2+z^3+\cdots## useful.
     
  6. Jan 2, 2017 #5

    vela

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    Yeah, but who remembers that? :wink:
     
  7. Jan 2, 2017 #6

    fresh_42

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    Wiki does. The advantage is, you don't even need a pencil.:smile:
     
  8. Jan 2, 2017 #7
    So taylor for ##\sin(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} ...## and for ##\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}+...##
    So:
    ##L=\lim_{x \rightarrow 0} \frac{1}{x^2}\left(\frac{(x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} ...)-x(1 - \frac{x^2}{2!} + \frac{x^4}{4!}+...)}{x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} ...}\right)##
    So beside the x getting canceled in the upper part what else can i do?
     
  9. Jan 2, 2017 #8

    fresh_42

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    Do what @Orodruin has said: calculate the difference, multiply with ##\frac{1}{x^2}##, cancel the quotient such that leading term of the denominator equals ##1## and see which terms without an ##x## are left. (I haven't done it this way, though.)
     
  10. Jan 2, 2017 #9
    Since the terms after ##\frac{x^3}{3!}## are getting smaller and smaller can i ignore them and write just the two leading members of the series?
    Like:
    ##L=\lim_{x \rightarrow 0} \frac{1}{x^2}\left(\frac{(x - \dfrac{x^3}{3!}) -x(1 - \frac{x^2}{2!})}{x - \dfrac{x^3}{3!}}\right)##
    ##L=\lim_{x \rightarrow 0}\left(\frac{(x - \dfrac{x^3}{3!}) -x(1 - \frac{x^2}{2!})}{x^3}\right)##
    ##L=\lim_{x \rightarrow 0}\left(\frac{x^3}{3x^3}\right)=\frac{1}{3}##
    Would this be correct?
     
  11. Jan 2, 2017 #10

    fresh_42

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    Yes, it is correct. Mathematically one would have notated the missing terms like this: ##\; \sin(x)=x-\frac{x^3}{3!}+O(x^5)## instead of just dropping them, but the result is the same.

    (##O(x^5)## means basically: "something with ##x^5## as factor".)
     
  12. Jan 2, 2017 #11

    Ray Vickson

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    This is not quite right, and can sometimes lead to an incorrect answer (as it does in this example). If we keep only ##1-x^2/2## from ##\cos x## and just retain ##x## from ##\sin x## we get
    $$\text{function} \doteq \frac{1}{x} \left( \frac{1}{x} - \frac{1}{x} + \frac{x^2}{2x} \right) = \frac{1}{2}.$$
    This would lead to the incorrect limit 1/2.

    However, if we keep ##x - x^3/6## from ##\sin x## we get
    $$ \text{function} \doteq \frac{1}{x} \left( \frac{1}{x} - \frac{1 - x^2/2}{x - x^3/6} \right) = \frac{1}{3} + O(x^2),$$
    giving the correct limit 1/3.

    When we have cancellations (as in ##1/x - \cot x##) we should keep a couple of additional terms just to be sure that all the cancellations are properly accounted for, and so we get a correct constant term.
     
    Last edited: Jan 3, 2017
  13. Jan 2, 2017 #12

    Orodruin

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    You misread me. I was referring to the expressions the OP had constructed with a common denominator. Clearly this does lead to the correct limit - as found by the OP.

    If you want to be sure about it you should not be throwing terms at all but instead insert O(x^3). This will immediately tell you if you missed a term of a relevant order.
     
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