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How do I find the limit of x as it approaches zero

  1. Jun 8, 2004 #1
    How do I find the limit of x as it approaches zero of sin(3x)/x? I seriously have no clue what to do here
     
  2. jcsd
  3. Jun 8, 2004 #2

    arildno

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    Set 3x=u, so that sin(3x)/x=3*sin(u)/u
     
  4. Jun 8, 2004 #3
    3sin(u)/u is the limit?
     
  5. Jun 8, 2004 #4
    [tex]\lim_{u \to 0} 3 \frac{\sin{u}}{u} = 3\lim_{u \to 0} \frac{\sin{u}}{u} = 3 \cdot 1 = 3[/tex]

    cookiemonster
     
  6. Jun 8, 2004 #5
    how did the three get placed outside the ()?

    How do I find the limit of x as it approaches zero of [1-sqrt(x^2+1)]/x^2?

    I sub in u so it becomes [1-sqrt(u^2+1)]/u^2...?
     
  7. Jun 8, 2004 #6

    arildno

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    Okay:
    [tex]\frac{\sin(3x)}{x}=3(\frac{\sin(3x)}{3x})[/tex]

    We now set u(x)=3x, and consider the function G(x)=F(u(x)), where:
    [tex]F(u)=3\frac{sin(u)}{u}[/tex]

    We are interested in computing [tex]\lim_{x\to{0}}G(x)[/tex]

    For a composite function with continous kernel u and "shell" F at [tex]x=x_{0}[/tex], we have:
    [tex]\lim_{x\to{x}_{0}}F(u(x))=F(\lim_{x\to{x}_{0}}u(x))=\lim_{u\to{u}_{0}}F(u), u_{0}=u(x_{0})[/tex]

    The limit of [tex]\frac{\sin{u}}{u}[/tex] as u goes to zero, is 1, as the Muffinangel states.
     
  8. Jun 8, 2004 #7
    Wait a moment, how exactly is
    [tex]\lim_{u \to 0}\frac{\sin u}{u} = 1[/tex]​
    I can see how this is true from a graphical perspective (if I graph sin u and u and look at their behaviour close to zero, or if I draw an appropriate triangle and see how it behaves as one of the angles goes to zero), but is there any ways of determining this algebraically.
     
  9. Jun 8, 2004 #8
    L'Hospital's rule.

    [tex]\lim_{x \to 0} \frac{\sin{x}}{x} = \lim_{x \to 0} \frac{\cos{x}}{1} = 1[/tex]

    Urban's second problem can be solved by L'Hospital's, as well.

    cookiemonster

    Edit: Who would'a thunk it... There's an s in L'Hospital!
     
    Last edited: Jun 8, 2004
  10. Jun 8, 2004 #9
    That is cheating since you need to now the derivative of sin x to use L'Hopital's rule and to compute the derivative of sin x you need to figure out that limit. You can't use taylor series either since that uses derivates as well. I need a more fundamental explanation.
     
  11. Jun 8, 2004 #10

    AKG

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    For one, you could use l'hopital's rule, but I think there should be a way to do it without that.

    The proof from my book essentially states that as we approach zero, [itex]\theta < \tan \theta[/itex] and that [itex]\frac{\sin \theta}{\theta} < 1[/itex], so we get [itex]\cos \theta < \frac{\sin \theta}{\theta} < 1[/itex] and by the Squeeze Theorem the limit is proved.
     
  12. Jun 8, 2004 #11
    All you need to know is that the derivative of sin(x) at 0 is 1. You don't need the actual derivative, no?

    cookiemonster
     
  13. Jun 9, 2004 #12

    arildno

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    "Proof" of the limit [tex]\lim_{\theta\to0}\frac{\sin\theta}{\theta}[/tex]

    Consider the three regions in the plane:
    a) A (unit-) circular section with angle [tex]\theta[/tex]
    b) The inscribed right-angle triangle with sides [tex]\sin\theta,\cos\theta,1[/tex]
    c) The "outer" triangle with short sides given by [tex]1,\tan\theta[/tex]

    Obviously, the respective areas A, B, C fulfill B<=A<=C
    So, we have, using the radian angle measure:
    [tex]\frac{1}{2}\sin\theta\cos\theta\leq\frac{1}{2}\theta\leq\frac{1}{2}\tan\theta[/tex]

    Or:
    [tex]\cos\theta\leq\frac{\theta}{\sin\theta}\leq\frac{1}{\cos\theta}[/tex]

    Taking the limit to zero angle yields by the Squeeze theorem, as AKG says, the result.
     
  14. Jun 9, 2004 #13

    HallsofIvy

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    "How do I find the limit of x as it approaches zero of [1-sqrt(x^2+1)]/x^2?"

    Multiply numerator and denominator by 1+ sqrt(x^2+1) and see what happens.
     
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