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Need help evaluating a limit

  1. May 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Evaluate the limit ##\lim_{x\to0} \frac{1}{x}(\frac{1}{tanx}-\frac{1}{x}) ## using Taylor's formula. (Hint: ##\frac{1}{1+c}=\frac{1-c^2+c^2}{1+c} ## may be useful)

    3. The attempt at a solution
    I began by substituting ##tanx## with ##x+\frac{x^3}{3}+x^3ε(x)##, where ε tends to zero as x approaches 0.

    ##\frac{1}{x}(\frac{1}{tanx}-\frac{1}{x})=\frac{1}{x}(\frac{x-tanx}{xtanx})=\frac{1}{x}(\frac{x-x-\frac{x^3}{3}-x^3ε(x))}{x(x+\frac{x^3}{3}+x^3ε(x))})=\frac{1}{x}(\frac{-\frac{x^3}{3}-x^3ε(x))}{x(x+\frac{x^3}{3}+x^3ε(x))})=\frac{-\frac{x^3}{3}-x^3ε(x))}{x^2(x+\frac{x^3}{3}+x^3ε(x))}=\frac{-\frac{x^3}{3}-x^3ε(x))}{x^3+\frac{x^4}{3}+x^4ε(x))}=\frac{-\frac{1}{3}-ε(x))}{1+\frac{x}{3}+xε(x))}## →-1/3 as x→0
     
    Last edited: May 3, 2016
  2. jcsd
  3. May 3, 2016 #2

    stevendaryl

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    You're not far from the answer. After cancellations, you have: [itex]\frac{\frac{x^3}{3} + ...}{x^3 + ...}[/itex], where [itex]...[/itex] represents higher-order terms. If you ignore the higher-order terms, what do you get?
     
  4. May 3, 2016 #3
    It's 1/3. But where to use the hint I am given?
     
    Last edited: May 3, 2016
  5. May 3, 2016 #4

    Samy_A

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    You lost a minus sign in the numerator.
     
  6. May 3, 2016 #5

    stevendaryl

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    I don't get that, either.
     
  7. May 3, 2016 #6
    You are right. It's a typo.
     
  8. May 3, 2016 #7

    PeroK

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    I think you've got the wrong series for ##tan(x)##. Check your coefficients. Also, if you are going to use the Taylor series, you should use the series for ##1/tan(x)## by applying the binomial expansion to the series for ##tan(x)## or using the series for ##cot(x)##.

    That said, this one looks tailor-made(!) for L'Hopital, using ##tan = sin/cos##.
     
  9. May 3, 2016 #8

    stevendaryl

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    No, I think he started with the wrong expansion for tan(x). It should be [itex]tan(x) = x + \frac{x^3}{3} + ...[/itex] not [itex]x - \frac{x^3}{3}[/itex]
     
  10. May 3, 2016 #9

    Samy_A

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    Yes, he has a wrong Taylor series. In the denominator, that should have been ##+\frac{x^3}{3}##. But in the numerator, he expands ##x-\tan x##, so the ##\frac{x^3}{3}## gets a minus sign.
    When taking the limit, the ##x³## term resulting from the ##\tan## series in the denominator is not important, but the one in the numerator is.
    The correct limit is ##-\frac13##.
     
  11. May 3, 2016 #10

    PeroK

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    My advice would be to do it the easy way using L'Hopital so you know what the answer is, then do it the hard way using Taylor series :wink:
     
  12. May 3, 2016 #11

    Samy_A

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    Sure, the hints he got are strange. This is, as you said, tailor-made(!) for L'Hopital.
     
  13. May 3, 2016 #12
    ##\frac{1}{x}(\frac{1}{tanx}-\frac{1}{x})=\frac{1}{x}(\frac{x-tanx}{xtanx})=\frac{1}{x}(\frac{x-x-\frac{x^3}{3}-x^3ε(x))}{x(x+\frac{x^3}{3}+x^3ε(x))})=...=\frac{-\frac{x^3}{3}-x^3ε(x))}{x^3+\frac{x^4}{3}+x^4ε(x))}=\frac{-\frac{1}{3}-ε(x))}{1+\frac{x}{3}+xε(x))}## →-1/3 as x→0

    Looks like I didn't need the hint. However, I think I'm supposed to derive the Taylor formula I used for the problem. We have covered expansions for sinx and cos x in class. So ##tanx=\frac{sinx}{cosx}=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}{1-\frac{x^2}{2!}+\frac{x^4}{4!}-...}=##
     
    Last edited: May 3, 2016
  14. May 3, 2016 #13

    PeroK

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    I would go for: ##1/tanx = cosx/sinx##

    You might as well make things easy for yourself!

    Use the binomial for ##1/sinx = (x-\frac{x^3}{3!}+\frac{x^5}{5!}- \dots)^{-1} = (1/x)(1-\frac{x^2}{3!}+\frac{x^4}{5!}- \dots)^{-1}##
     
  15. May 5, 2016 #14
    I actually derived the expansion for ##\tan x## from the definition of taylor formula. ##f'(x)=D\tan x=1+\tan^2 x##, ##f'(0)=1##
    ##f''(x)=2\tan x(1+\tan^2 x)##, ##f''(0)=0## etc. and got ##\tan x=x+\frac{x^3}{3}+x^3ε(x)##. I think this is maybe shorter and nicer than deriving it by long division.
     
    Last edited: May 5, 2016
  16. May 5, 2016 #15

    SammyS

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    I thought I had posted something on this a day or two ago. Using the first of Perok's suggestions, then common denominator, etc. :

    ##\displaystyle \
    \frac{1}{x}\left(\frac{1}{\tan x}-\frac{1}{x}\right)
    \ ##

    ##\displaystyle \
    =\frac{x \cos x - \sin x}{x^2\sin x}
    \ ##

    Now use the Taylor expansions for sin x and cos x .
     
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