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Lim -> inf, x(sin(1/x)) = 1, why?

  1. Apr 8, 2005 #1
    I would've thought

    [tex] \lim_{x \rightarrow \infty} x sin(\frac{1}{x})} = 0 [/tex]

    because

    [tex] \lim_{x \rightarrow \infty} x = \infty [/tex] and [tex]\lim_{x \rightarrow \infty} sin(\frac{1}{x})} = sin ( \lim_{x \rightarrow \infty} \frac{1}{x} = sin( 0)= 0[/tex] and [tex] \infty * 0 = 0 [/tex]

    I begin to wonder if they should go back to teaching infestimals because in cases
     
  2. jcsd
  3. Apr 8, 2005 #2

    Imo

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    I'll quickly point out...zero times infinity is not always zero. Take for instance
    [tex] \lim_{x \rightarrow \infty} \frac{x}{x}} = 1 [/tex]
    but before simplification it's
    [tex] \infty * 0 [/tex]
     
  4. Apr 8, 2005 #3

    Galileo

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    You are only allowed to split the limit into several limits (products, sums etc) if the individual limits exist. Clearly if one of them doesn't exist (infinity), you are no longer allowed to do so. Thus

    [tex]\lim_{x \to a}(f(x)+g(x))=\lim_{x \to a}f(x)+\lim_{x \to a}g(x)[/tex]
    and
    [tex]\lim_{x \to a}f(x)g(x)=\lim_{x \to a}f(x)\cdot \lim_{x \to a}g(x)[/tex]
    hold only if the limits for f(x) and g(x) exist.

    To solve your limit, try to find a link with:

    [tex]\lim_{x \to 0}\frac{\sin(x)}{x}[/tex]
     
  5. Apr 8, 2005 #4
    Can i mathematically write? :

    [tex]\lim_{x \to 0} (\frac{\sin(x)}{x}) = [/tex]

    [tex] (\frac{\sin(x \to 0)}{x \to 0}) = [/tex]

    given that [tex] sin(x \to 0)[/tex] itself approaches [tex](x \to 0) [/tex] as [tex] (x \to 0)[/tex],

    [tex] (\frac{x \to 0}{x \to 0}) = 1[/tex]

    here I am replacing x with the symbol [tex] x \to 0[/tex] and using it as a algebraic variable.
     
    Last edited: Apr 8, 2005
  6. Apr 8, 2005 #5

    dextercioby

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    Alebragic variables are a special kind of variables & need to be manipulated with care.
    The answer is still 1,though...

    Daniel.
     
  7. Apr 8, 2005 #6
    er so that's a yes but be careful how i manipulate them? treating x in this way makes some limits much easier to evaluate.
     
  8. Apr 8, 2005 #7

    dextercioby

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    We have a thread especially on this "sinc" limit...It's around here.

    Daniel.
     
  9. Apr 8, 2005 #8
    what thread are you referring to? a search on sinc limit only pulls up this thread.
     
  10. Apr 8, 2005 #9

    dextercioby

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  11. Apr 8, 2005 #10

    Gokul43201

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    What happens if you substitute y in the place of 1/x in the original problem ?
     
  12. Apr 8, 2005 #11
    No. Consider

    [tex]\lim_{x\rightarrow 0} \frac{1-\cos x}{x}[/tex]
     
  13. Apr 8, 2005 #12

    dextercioby

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    What Data meant was that [itex] 1-\cos x [/itex] approaches zero as [itex] x^{2} [/itex] (hence "faster") than [itex]x [/itex]...

    Daniel.
     
  14. Apr 8, 2005 #13
    l'hopital's rule can be applied to the limit. all you have to do is manipulate the problem so that you get a [itex]\frac{0}{0}[/itex] or [itex]\frac{\infty}{\infty}[/itex].

    [tex]\lim_{x \to \infty}xsin(\frac{1}{x})=\lim_{x \to \infty} \frac{sin( \frac{1}{x})}{\frac{1}{x}}=\frac{0}{0}[/tex]

    now, you should be able to apply l'hopital's rule and find why it is 1.
     
  15. Apr 8, 2005 #14

    xanthym

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    ktpr2:
    Are you familiar with l'Hospital's Rule?? This rule applies to all "0/0" cases mentioned here so far. In a certain sense, the rule involves the "speed" with which the numerator and denominator approach 0 because the Rule uses derivatives to resolve the indeterminate "0/0" forms. (It can also be applied similarly to "∞/∞" forms.) Thus:


    [tex] 1: \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}}\color{black} \, \frac {\sin(x)} {x} \ \ = \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}} \color{black} \, \frac{ \displaystyle \left (\frac {d} {dx} \sin(x) \right )} {\displaystyle \left ( \frac {d} {dx} \left( x \right) \right )} \color{black}} } \ \ = \ \ \frac { \cos \color{blue} \mathbf{ \left( 0 \right) }} {\color{blue} \mathbf{ \left( 1 \right) } } \ \ = \ \ \color{red} \mathbf{ \left( 1 \right) } [/tex]


    [tex] 2: \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}}\color{black} \, \frac {1 \, - \, \cos(x)} {x} \ \ = \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}}\color{black} \, \frac{ \displaystyle \left (\frac {d} {dx} \left \{ 1 \, - \, \cos(x) \right \} \right )} {\displaystyle \left ( \frac {d} {dx} \left( x \right) \right ) } \ \ = \ \ \frac { \sin \color{blue} \mathbf{ \left( 0 \right) }} {\color{blue} \mathbf{ \left( 1 \right) } } \ \ = \ \ \color{red} \mathbf{ \left( 0 \right) } [/tex]



    ~~
     
    Last edited: Apr 9, 2005
  16. Apr 8, 2005 #15

    dextercioby

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    It's logically incorrect to apply l'Hôpital's rule to those 2 limits...:wink: Read the thread i made a link to in post #9.In that thread u'll find another link to an illuminating wikipedia page.

    Daniel.
     
  17. Apr 8, 2005 #16

    dextercioby

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    Shame you used such a pretty writing for something incorrect.

    Daniel.
     
  18. Apr 8, 2005 #17

    xanthym

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    Minor surgery performed to repair the previous hasty logical leap:


    [tex] 1: \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}}\color{black} \, \frac {\sin(x)} {x} \ \ = \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}} \color{black} \, \frac{ \displaystyle \left (\frac {d} {dx} \sin(x) \right )} {\displaystyle \left ( \frac {d} {dx} \left( x \right) \right )} \color{black}} } \ \ = \ \ \frac { \cos \color{blue} \mathbf{ \left( 0 \right) }} {\color{blue} \mathbf{ \left( 1 \right) } } \ \ = \ \ \color{red} \mathbf{ \left( 1 \right) } [/tex]


    [tex] 2: \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}}\color{black} \, \frac {1 \, - \, \cos(x)} {x} \ \ = \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}}\color{black} \, \frac{ \displaystyle \left (\frac {d} {dx} \left \{ 1 \, - \, \cos(x) \right \} \right )} {\displaystyle \left ( \frac {d} {dx} \left( x \right) \right ) } \ \ = \ \ \frac { \sin \color{blue} \mathbf{ \left( 0 \right) }} {\color{blue} \mathbf{ \left( 1 \right) } } \ \ = \ \ \color{red} \mathbf{ \left( 0 \right) } [/tex]


    ~~
     
    Last edited: Apr 9, 2005
  19. Apr 8, 2005 #18

    dextercioby

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    U didn't read the thread,nor the wiki page,did u?

    U can't use the derivative of "sine" & "cosine",unless u adopt another method for computing them,other than the definition...

    Daniel.
     
  20. Apr 8, 2005 #19

    xanthym

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    Nobody is trying to "prove" anything here (Msgs #14 & #17).
    The l'Hospital's Rule is being offered as an efficient method for determining limits of indeterminate forms.


    ~~
     
  21. Apr 8, 2005 #20

    dextercioby

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    How do you prove that

    [tex] \frac{d}{dx}\sin x=\cos x [/tex]

    ?

    Daniel.
     
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