# Cosx math problem

Am i doing these right?

lim x->0 (1-cosx) all over x
=(1-cosx/x)(1+cosx/1+cosx)
=(1-cos^2x)/x(1+cosx)
=(sin^2x)/x(1+cosx)
=x(sin^2x)/xx(1+cosx)
=0/2
=0

are my steps right?

Nevermind, thats indeterminate. Your denominator is still 0. You may have to use l'hopitals rule.

AKG
Homework Helper
=x(sin^2x)/xx(1+cosx)
=0/2

That step is not right. The top will be 0*(sin²0) which is 0, the bottom will be 0 * 0 * 2 which is 0, not 2.

What you can do is recognize that:

x(sin²x)/x²(1 + cosx)
= [x/(1 + cosx)][sin²x/x²]
= [x/(1 + cosx)](sinx/x)²

The limit as x -> 0 of (sinx/x) exists, and you should know it, and the limit as x -> 0 of [x/(1 + cosx)] exists, and you should be able to easily compute that, so the limit of the product is the product of the limits. That way, you get (0/2)(1)². Perhaps this is how you got rid of the x² on the bottom, and hence went from

=x(sin^2x)/xx(1+cosx)

to

=0/2

but it's not clear that you did that. At least show your work. Also, make sure that when you write it, don't just drop the "lim x->0" part until you actually compute it.

HallsofIvy
Homework Helper
$$\frac{1-cos(x)}{x}\frac{1+cos(x)}{1+cos(x)}$$
$$=\frac{sin^2(x)}{x(1+cos(x))}$$
$$=\frac{sin(x)}{x}\frac{sin(x)}{1+cos(x)}$$

What exactly do you already know and can use?
If you are allowed to use the well known limit: $$\frac{sin x}{x}$$ goes to 1 and the facts that sine and cosine are continuous, so that the second fraction goes to 0/2= 0, it follows that the limit is 0.

Galileo
Homework Helper
I see no reason why the steps are inorect. Gillgill probably multiplied top and bottom by x, in order to write:

$$\frac{\sin^2 x}{x(1+cosx)}=\frac{x\sin^2 x}{x^2(1+cosx)}=\left(\frac{\sin x}{x}\right)^2\frac{x}{1+\cos x}$$

so you can use that: $\frac{\sin x}{x}$ goes to one and all three products of limits exist.

$$\lim _{x \to 0}\left(\frac{\sin x}{x}\right)^2\frac{x}{1+\cos x}=(1)(1)(0)=0$$

EDIT: It doesn't show in the calculations though, nevermind.

Last edited:
dextercioby