Without using L' Hospital's rule,how to get 0 as the answer for limit of (1cos 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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 14032013 09:55

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 14032013 10:38
(Original post by MR INFINITY)
Without using L' Hospital's rule,how to get 0 as the answer for limit of (1cos 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.
Then you can use .Last edited by Notnek; 14032013 at 10:42. 
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 14032013 10:52
(Original post by MR INFINITY)
Without using L' Hospital's rule,how to get 0 as the answer for limit of (1cos 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.
Last edited by Farhan.Hanif93; 14032013 at 10:54. 
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 14032013 10:59
(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).
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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 14032013 11:06
(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. 
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 14032013 11:10
[QUOTE=Farhan.Hanif93;41804455]Observe that (think carefully about how you'd show the RH inequality).
Could you explain how you get 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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 14032013 11:31
(Original post by MR INFINITY)
Could you explain how you get and what RH inequality is? I'm only one more step to squeeze theorem and get 0 as the answer.
Thanks.
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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 14032013 12:29
(Original post by Farhan.Hanif93)
The LH inequality arises from the boundedness of cosx
(Original post by Farhan.Hanif93)
RH = RH inequality arises from the power series definition of cosx.
cos x = 1  0  (x^2)/4 + 0 + ...
cos x =1 (x^2)/4 + ...
cos x ≥ 1  (x^2)/4
1cos x≤ (x^2)/4
1cos 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.Last edited by MR INFINITY; 14032013 at 12:30. 
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 14032013 13:04
(Original post by MR INFINITY)
From cos x ≤ 1,I get 0 ≤ 1cos x.
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
1cos x≤ (x^2)/4
1cos x≤ x^2
Plus 2!≠4.
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.
EDIT: Also, is this actually something to do with ALevels or did you mislabel the thread? 
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 14032013 13:20
(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.
Unit circle:
What are the lengths of the arc AB and the line AB? 
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 14032013 13:58
(Original post by Farhan.Hanif93)
Instead, you define cos x by the power series and build your argument from there.
(Original post by Farhan.Hanif93)
Plus 2!≠4. 
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 14032013 14:11
(Original post by notnek)
What are the lengths of the arc AB and the line AB?
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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 14032013 14:27
(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?
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

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 14032013 15:11
(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
http://math.stackexchange.com/questi...queezetheorem
Is this what they mean?
<x approaches zero)lim(1cos x)/x
<x approaches zero)lim (1cos^2 x)/(x+cos x)
(1square of cos 0) / (0+1)
(11)/(0+1)
so the limit =0 
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 15032013 04:39
(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
Otherwise, the geometrical approach works fine.
PS: since you're trying to prove that you've probably shown that as
You could then take the following shortcut by whipping out a trig identity:
... and avoid the limit altogether! 
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 15032013 21:44
(Original post by MR INFINITY)
Still can't figure things out.My geometry is bad enough.
http://math.stackexchange.com/questi...queezetheorem
Is this what they mean?
<x approaches zero)lim(1cos x)/x
<x approaches zero)lim (1cos^2 x)/(x+cos x)
(1square of cos 0) / (0+1)
(11)/(0+1)
so the limit =0
The limit of is a
wellknown question to use the
Because
And why ?
Look at the picture bellow:
For the area (A) of triangles and Section
that is
using x>0 as sinx/x is even, and cosx>1 as x>0 from the sandwich rule
follows
Last edited by ztibor; 15032013 at 21:47.
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