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Sinx/x x tends to 0

  1. Jan 28, 2009 #1
    I know that as x tends to 0, (sinx)/x tends to 1. A post from GibZ got me thinking, would this be a proper proof of that:
    [tex]\lim_{x \rightarrow 0} x = \lim_{x \rightarrow 0} sinx[/tex] and so [tex]\lim_{x \rightarrow 0} \frac{x}{x} = 1 = \lim_{x \rightarrow 0} \frac{sinx}{x}[/tex] ??
     
  2. jcsd
  3. Jan 28, 2009 #2
    Re: Sinx/x

    No! Think about using your method to find
    [tex]\lim_{x \to 0} x[/tex] by writing [tex]x=\frac{ x^2 }{ x }[/tex]. Also the limit of a quotient is only the quotient of the limits when the bottom limit is nonzero!

    Use L'Hospital!!!
     
  4. Jan 28, 2009 #3
    Re: Sinx/x

    Ah! That's the part I forgot! That the limit of a quotient is a quotient of limits only when the denominator isn't 0. I constructed a counter-example too to prove myself wrong in the previous post.

    [tex]\lim_{x \rightarrow 0} x = 0 [/tex] and so [tex]\lim_{x \rightarrow 0} \frac{x}{x} = 1 = \lim_{x \rightarrow 0} \frac{0}{x} = 0 [/tex] which obviously isn't true (because the denominator is tending to 0).

    Thanks for the reply :)
     
  5. Jan 29, 2009 #4

    lurflurf

    User Avatar
    Homework Helper

    Re: Sinx/x


    Why do people want to use L'Hospital on every problem? Is it because it is fun to say? L'Hospital is not needed.
    lim sin(x)/x=lim [sin(0+x)-sin(x)]/[(0+x)-0]=sin'(0)=1

    We might want to compute the limit directly from the definition. For that we need to choose a definition.

    def1:
    sin(x+y)=sin(x)cos(y)+cos(x)sin(y)
    cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
    cos(x)^2+sin(x)^2=1
    lim sin(x)/x=0

    here the limit is included in the definition
    QED

    def2:
    sin(x)=x-x^3/6+...

    we can write
    sin(x)=sin(x)+sin'(0) x+sin''(0) x^2/2+sin'''(0) x^3/6+...
    equating like terms yields
    sin'(0)=1
    QED

    def3:
    sin''(x)+sin(x)=0
    sin(0)=0
    sin'(0)=1

    sin'(0)=1 is in the definition
    QED

    def4:
    sin(x)/x=(1-[x/(1pi)]^2)(1-[x/(2pi)]^2)(1-[x/(3pi)]^2)(1-[x/(4pi)]^2)(1-[x/(5pi)]^2)...

    lim sin(x)/x=(1-[0/(1pi)]^2)(1-[0/(2pi)]^2)(1-[0/(3pi)]^2)(1-[0/(4pi)]^2)(1-[0/(5pi)]^2)...
    =1
    QED

    def5:
    various geometric junk from which it is noticed

    for x small
    sin(2x)/(2x)<=cos(x)<=sin(x)/x<=1
    from which we conclude
    cos(0)<=lim sin(x)/x<=1
    lim sin(x)/x=1
    QED
     
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