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Limx->0 (sinx/x)=1?

  1. Jun 11, 2012 #1
    limx-->0 (sinx/x)=1?

    1. The problem statement, all variables and given/known data

    Hi I have the question evaluate the limit going to zero of

    (cos(x)-1)/sin(x2)



    2. Relevant equations



    3. The attempt at a solution

    lim x→0 (cos(x)-1)/sin(x2)

    goes to 0/0 by Lhoptal's

    f'(x)/g'(x)

    = (limx→0 sinx/x)(limx→0 -1/ 2cos(x2))

    but the answer I have says that (limx→0 sinx/x) = 1

    how is this so

    I know the other term goes to -1/2 but yeah I'm really confused about the sin term going to 1 is this a mistake or is there something i don't know?

    thanks.
     
  2. jcsd
  3. Jun 11, 2012 #2
    Re: limx-->0 (sinx/x)=1?

    Hint:
    [tex]
    \cos x - 1 = -2 \sin^2 \left( \frac{x}{2} \right)
    [/tex]
    Divide the numerator and denominator by [itex]x^2[/itex], and use your limit.
     
  4. Jun 11, 2012 #3
    Re: limx-->0 (sinx/x)=1?

    do you mean cos2x-1=-2sin2(x/2)?

    if i divide numerator and denominator by x2

    (sinx × x2)/x3

    I still get 0 out of it when I plug 0 into anything times by sin i get 0.

    do I take L'hopitals again?

    of the individual sin part of the function? can you partially diff and apply L'hopitals?
     
  5. Jun 11, 2012 #4
    Re: limx-->0 (sinx/x)=1?

    No, because:
    a) it is an incorrect trigonometric identity;
    b) it is not useful for your problem, since the left-hand side is not what you have in your limit.

    I meant exactly what I wrote:
    [tex]
    \cos x - 1 = -2 \, \sin^2 \left( \frac{x}{2} \right)
    [/tex]

    I don't know how you got this.
     
  6. Jun 11, 2012 #5
    Re: limx-->0 (sinx/x)=1?

    What happens if you try L'Hopital again?
     
  7. Jun 11, 2012 #6

    HallsofIvy

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    Re: limx-->0 (sinx/x)=1?

    I think I would be inclined to write it as
    [tex]\frac{cos(x)- 1}{sin(x^2)}= \frac{x^2}{sin(x^2)}\frac{cos(x)- 1}{x^2}[/tex]

    It's easy to see that [itex]sin(x^2)/x^2[/itex] goes to 1. Using L'Hopital's rule on [tex]\frac{cos(x)- 1}{x^2}[/tex] leads to [tex]\frac{-sin(x)}{2x}[/tex] which has limit -2.
     
  8. Jun 11, 2012 #7
    Re: limx-->0 (sinx/x)=1?

    Ok, I can see how you get that but the part I'm confused with is that you have a cos factor in the numerator and a sin factor in the denominator. every time L'hopital's gets used these just swap from sin to cos visca versa.

    and cos 0 =1 but sin0=0

    thats how I don't understand how the zero term vanishes?

    can you use L'hopitals rule on just part of a fraction?

    because it looks like every time I differentiate tops and bottoms I have a sin and a cos factor, when i really only can have cos factors.?
     
  9. Jun 11, 2012 #8
    Re: limx-->0 (sinx/x)=1?

    This is much more easily solved with Taylor expansion.

    [tex]\cos x - 1 = \left(1 - \frac{x^2}{2} + \ldots{}\right) - 1 = -\frac{x^2}{2} + \ldots{}[/tex]

    And for the denominator,

    [tex]\sin x^2 = x^2 - \frac{x^6}{6} + \ldots{}[/tex]

    So the limit can be written as

    [tex]\lim_{x \to 0} \frac{\cos x - 1}{\sin x^2} = \lim_{x\to 0} \frac{-x^2/2 + \ldots}{x^2 - x^6/6 + \ldots{}}[/tex]

    The limit is as [itex]x \to 0[/itex], so throw away all but the lowest power term in both the numerator and denominator, and you clearly get

    [tex]\lim_{x \to 0} \frac{-x^2/2}{x^2} = -\frac{1}{2}[/tex]

    Now, you originally asked how it could be that,

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

    Again, this is easily solved by Taylor expansion.

    [tex]\lim_{x \to 0} \frac{x - x^3/6 + \ldots}{x} = \lim_{x\to 0} \frac{x}{x} = 1[/tex]

    Or by l'Hopital's rule,

    [tex]\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1[/tex]
     
  10. Jun 11, 2012 #9

    HallsofIvy

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    Re: limx-->0 (sinx/x)=1?

    I'm not sure what you mean by that. "vanishes" means "is 0".

    You understand what L'Hopital's rule is, dont you?
    [tex]\lim_{x\to a}\frac{f(x)}{g(x)}= \frac{\lim_{x\to a}f'(x)}{\lim_{x\to a} g'(x)}[/tex]
    provided those limits exist. I dont know what you mean by "part of a fraction". If you can write a fraction as a sum or product, then you can use [itex]\lim_{x\to a} (f+g)(x)= \lim_{x\to a}f(x)+ \lim_{x\to a}g(x)[/itex] and [itex]\lim_{x\to a}fg(x)= \left(\lim_{x\to a}f(x)\right)\left(\lim_{x\to a}g(x)\right)[/itex].

    Again, I really don't know what you mean by this. Why would you think "I really only can have cos factors"?
     
  11. Jun 11, 2012 #10

    vela

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    Re: limx-->0 (sinx/x)=1?

    This method likely requires material the OP doesn't yet know considering he or she is still learning how to evaluate basic limits.
     
  12. Jun 11, 2012 #11
    Re: limx-->0 (sinx/x)=1?

    yep.
     
  13. Jun 11, 2012 #12
    Re: limx-->0 (sinx/x)=1?

    ok so its ok to have...

    limx→0 (-1/2cos(x2)limx→0(cos(x)/1)

    = -1/2 (1) =-1/2

    ??
     
  14. Jun 11, 2012 #13

    vela

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    Re: limx-->0 (sinx/x)=1?

    Applying L'Hopital's rule to what you started with, you got
    $$\lim_{x \to 0} \frac{\cos x - 1}{\sin x^2} = \lim_{x \to 0} \frac{-\sin x}{2x\cos x^2}$$ I think what you're missing is this next step. What happened was they broke the single limit into a product of two limits:
    $$\lim_{x \to 0} \frac{-\sin x}{2x\cos x^2} = \left(\lim_{x \to 0} \frac{\sin x}{x}\right)\left(\lim_{x \to 0} \frac{-1}{2\cos x^2}\right)$$ which you can do if the limits exist. So while it looked like L'Hopital's rule was only being applied to part of the fraction, namely, the sin x/x part, it really wasn't. It's just that the original limit was split into two separate limits. Now the first limit you can evaluate using L'Hopital's rule again.
     
  15. Jun 12, 2012 #14
    Re: limx-->0 (sinx/x)=1?

    Thanks vela that completely cleared it up.
     
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