Trigonometry limit

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.

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 $\displaystyle \lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$.

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.

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.

The LH inequality arises from the boundedness of cosx

RH = RH inequality arises from the power series definition of cosx.

From cos x ≤ 1,I get 0 ≤ 1-cos 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
1-cos x≤ (x^2)/4
1-cos x≤ x^2

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/CalculusLimitSinxOverxIs1.aspx )Thanks again.

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/CalculusLimitSinxOverxIs1.aspx )Thanks again.

Instead, you define cos x by the power series and build your argument from there.

Plus 2!≠4.

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?

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 $\displaystyle \frac{1-\cos x}{x}<\sin \frac{x}{2}$

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

Still can't figure things out.My geometry is bad enough.




http://math.stackexchange.com/questions/36299/finding-the-limit-of-1-cosx-x-as-x-to-0-with-squeeze-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