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    Without using L' Hospital's rule,how to get 0 as the answer for limit of (1-cos x)/x as x approaches 0?

    I did some research previously.I don't want to use L' Hospital's rule because I'm in the process of proving the derivative of sin x is cos x.

    Appreciate your response.
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    (Original post by MR INFINITY)
    Without using L' Hospital's rule,how to get 0 as the answer for limit of (1-cos x)/x as x approaches 0?

    I did some research previously.I don't want to use L' Hospital's rule because I'm in the process of proving the derivative of sin x is cos x.

    Appreciate your response.
    Try multiplying top and bottom by 1+\cos x.


    Then you can use \displaystyle \lim_{x\rightarrow 0} \frac{\sin x}{x} = 1.
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    (Original post by MR INFINITY)
    Without using L' Hospital's rule,how to get 0 as the answer for limit of (1-cos x)/x as x approaches 0?

    I did some research previously.I don't want to use L' Hospital's rule because I'm in the process of proving the derivative of sin x is cos x.

    Appreciate your response.
    Observe that 0\leq 1-\cos x \leq x^2 (think carefully about how you'd show the RH inequality). By considering the cases x>0 and x<0 separately, apply the squeeze theorem to get the result.


    (Original post by notnek)
    Try multiplying top and bottom by 1+\cos x.


    Then you can use \displaystyle \lim_{x\rightarrow 0} \frac{\sin x}{x} = 1.
    Given that both the limit you're trying to prove and the limit you're assuming are proven in essentially the same way, assuming one is just as big an assumption as assuming the other. You may as well prove the one we want, if you're going to assume that limit (which requires proof nonetheless).
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    (Original post by Farhan.Hanif93)
    Given that both the limit you're trying to prove and the limit you're assuming are proven in essentially the same way, assuming one is just as big an assumption as assuming the other. You may as well prove the one we want, if you're going to assume that limit (which requires proof nonetheless).
    Agreed.

    I'm a bit rusty and couldn't think of a way to prove it. I got too used to quoting sinx/x for trig limits.
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    (Original post by notnek)
    Agreed.

    I'm a bit rusty and couldn't think of a way to prove it. I got too used to quoting sinx/x for trig limits.
    Fair enough.
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    [QUOTE=Farhan.Hanif93;41804455]Observe that 0\leq 1-\cos x \leq x^2 (think carefully about how you'd show the RH inequality).

    Could you explain how you get 0\leq 1-\cos x \leq x^2 and what RH inequality is? I'm only one more step to squeeze theorem and get 0 as the answer.

    Thanks.
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    (Original post by MR INFINITY)
    Could you explain how you get 0\leq 1-\cos x \leq x^2 and what RH inequality is? I'm only one more step to squeeze theorem and get 0 as the answer.

    Thanks.
    RH = right hand so the RH inequality is 1-\cos x \leq x^2.

    The LH inequality arises from the boundedness of cosx, and the RH inequality arises from the power series definition of cosx. I'll leave you to fill in the details; please post back with some working if you want further help.
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    (Original post by Farhan.Hanif93)
    The LH inequality arises from the boundedness of cosx
    As x approaches 0,value of f(x)=cos x gets closer to 0.(considering only Quadrant I as we are looking on values very near 0). From cos x ≤ 1,I get 0 ≤ 1-cos x.

    (Original post by Farhan.Hanif93)
    RH = RH inequality arises from the power series definition of cosx.
    cos x =cos 0 - (sin0)(x) - (cos 0)*(x^2)/2! + (sin 0)*(x^3)/3!+...
    cos x = 1 - 0 - (x^2)/4 + 0 + ...
    cos x =1- (x^2)/4 + ...
    cos x ≥ 1 - (x^2)/4
    1-cos x≤ (x^2)/4
    1-cos x≤ x^2

    Now I get it.Thanks for your brilliant tips. It seems like this method requires formula d(sin x)/dx=cos x or d(cos x)/dx=- sin x (same reason I don't prefer L' Hospital' rule). Do you know of a method that does not involve the two formulas?(something like http://2000clicks.com/mathhelp/Calcu...xOverxIs1.aspx )Thanks again.
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    (Original post by MR INFINITY)
    From cos x ≤ 1,I get 0 ≤ 1-cos x.
    Yep

    cos x =cos 0 - (sin0)(x) - (cos 0)*(x^2)/2! + (sin 0)*(x^3)/3!+...
    cos x = 1 - 0 - (x^2)/4 + 0 + ...
    cos x =1- (x^2)/4 + ...
    cos x ≥ 1 - (x^2)/4
    1-cos x≤ (x^2)/4
    1-cos x≤ x^2
    Careful, I didn't say anything about using Maclaurin (as that assumes knowledge of the derivatives). Instead, you define cos x by the power series and build your argument from there.

    Plus 2!≠4. :p:

    It seems like this method requires formula d(sin x)/dx=cos x or d(cos x)/dx=- sin x (same reason I don't prefer L' Hospital' rule). Do you know of a method that does not involve the two formulas?(something like http://2000clicks.com/mathhelp/Calcu...xOverxIs1.aspx )Thanks again.
    It shouldn't (as I suggested in my previous comment). I can't say I know a geometric argument for a 'squeeze-theorem'-esque inequality in this case but it certainly wouldn't surprise me if there were one.

    EDIT: Also, is this actually something to do with A-Levels or did you mislabel the thread?
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    (Original post by MR INFINITY)
    Now I get it.Thanks for your brilliant tips. It seems like this method requires formula d(sin x)/dx=cos x or d(cos x)/dx=- sin x (same reason I don't prefer L' Hospital' rule). Do you know of a method that does not involve the two formulas?(something like http://2000clicks.com/mathhelp/Calcu...xOverxIs1.aspx )Thanks again.
    You can use a geometric proof to show that 1-\cos x&lt;\frac{x}{2}.

    Unit circle:



    What are the lengths of the arc AB and the line AB?
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    (Original post by Farhan.Hanif93)
    Instead, you define cos x by the power series and build your argument from there.
    How is the power series then? I haven't actually started my A level (the proper way at school) so I still don't know many things.

    (Original post by Farhan.Hanif93)
    Plus 2!≠4. :p:
    Yes. 2!= 2*1 which gives 2!= 2​ .
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    (Original post by notnek)
    What are the lengths of the arc AB and the line AB?
    Arc AB = 1*x = x

    By Pythagoras theorem,
    (Line AB)^2 = (sin x)^2 + (1+cos x)^2
    = (sin x)^2 + 1 - 2cos x +(cos x)^2
    Applying Pythagorean identity,
    = 1 + 1 - 2cos x
    (Line AB)^2 = 2 (1- cos x)

    Arc AB is longer than line AB.
    x ≥ 2 sqrt(1- cos x)
    x/2 ≥ sqrt(1- cos x)

    That sqrt is bothering me! Where did I get wrong?
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    (Original post by MR INFINITY)
    Arc AB = 1*x = x

    By Pythagoras theorem,
    (Line AB)^2 = (sin x)^2 + (1+cos x)^2
    = (sin x)^2 + 1 - 2cos x +(cos x)^2
    Applying Pythagorean identity,
    = 1 + 1 - 2cos x
    (Line AB)^2 = 2 (1- cos x)

    Arc AB is longer than line AB.
    x ≥ 2 sqrt(1- cos x)
    x/2 ≥ sqrt(1- cos x)

    That sqrt is bothering me! Where did I get wrong?
    Sorry that was my mistake. It should be

    \displaystyle 1-\cos x &lt; \frac{x^2}{4}


    Alternatively, if D is the point where the vertical from A meets OB then consider the angle DAB which is x/2 and you can show

    \displaystyle \frac{1-\cos x}{x}&lt;\sin \frac{x}{2}
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    (Original post by notnek)
    Alternatively, if D is the point where the vertical from A meets OB then consider the angle DAB which is x/2 and you can show \displaystyle \frac{1-\cos x}{x}&lt;\sin \frac{x}{2}
    Still can't figure things out.My geometry is bad enough.




    http://math.stackexchange.com/questi...queeze-theorem
    Is this what they mean?

    <x approaches zero)lim(1-cos x)/x
    <x approaches zero)lim (1-cos^2 x)/(x+cos x)
    (1-square of cos 0) / (0+1)
    (1-1)/(0+1)
    so the limit =0
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    (Original post by MR INFINITY)
    How is the power series then? I haven't actually started my A level (the proper way at school) so I still don't know many things
    The idea is that the cosine function, as many others, can be defined in several manners (a power series, a ratio, the solution to a particular differential equation...); usually at a certain level you will go with the former, because it is the most useful. So what you're allowed to use for this question really depends on what definition you're working with. If your starting point is the cosine's power series then, as Farhan suggested, you can proceed likewise:

    \cos x= 1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots\Rightarrow 0\leq 1-\cos x\leq \frac{1}{2}x^2

    Otherwise, the geometrical approach works fine.

    PS: since you're trying to prove that \sin'x = \cos you've probably shown that \frac{\sin x}{x}\to 1 as x\to 0

    You could then take the following shortcut by whipping out a trig identity:

    \displaystyle\dfrac{\sin (x+h)-\sin x}{h}=\frac{\sin(h/2)}{(h/2)}\cdot\cos\left(\frac{2x+h}{2} \right)

    ... and avoid the \cos limit altogether!
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    (Original post by MR INFINITY)
    Still can't figure things out.My geometry is bad enough.




    http://math.stackexchange.com/questi...queeze-theorem
    Is this what they mean?

    <x approaches zero)lim(1-cos x)/x
    <x approaches zero)lim (1-cos^2 x)/(x+cos x)
    (1-square of cos 0) / (0+1)
    (1-1)/(0+1)
    so the limit =0
    Why do you complicate the simple.
    The limit of \lim_{x\rightarrow 0}\frac{1-\cos x}{x} is a
    well-known question to use the \lim_{X \rightarrow 0} \frac{\sin X}{X}=1

    Because
    1-\cos x=2\sin^2\frac{x}{2}
    \frac{1-\cos x}{x}=\frac{x}{2}\cdot \left (\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right )^2

    And why \lim_{X \rightarrow 0} \frac{\sin X}{X}=1?
    Look at the picture bellow:
    Name:  limit.PNG
Views: 42
Size:  5.1 KB
    For the area (A) of triangles and Section
    A_{OAP \Delta} &lt;A_{OAP sec}&lt;A_{OAB \Delta}
    that is
    \frac{1\cdot \sin x}{2}&lt;\frac{1^2\cdot x}{2}&lt;\frac{1\cdot \tan x}{2}
    1&lt;\frac{x}{\sin x}&lt;\frac{1}{\cos x}
    1&gt;\frac{\sin x}{x}&gt;\cos x
    using x>0 as sinx/x is even, and cosx->1 as x->0 from the sandwich rule
    follows
    \lim_{x \rightarrow 0} \frac{\sin x}{x}=1
 
 
 
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