Limit $\lim_{x \to 0} \dfrac{\sin(1-\cos x)}{ x} $ $$\lim_{x \to 0} \dfrac{\sin(1-\cos x)}{ x} $$
What is wrong with this argument: as $x$ approaches zero, both $x$ and $(1-\cos x)$ approaches $0$.
So the limit is $1$ . 


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*How can we prove that they approaches zero at same rate?

*This is not about solving the limit because I already solved it but about the rate of both functions going to zero .
m referring to $$\lim_{x \to 0} \dfrac{\sin x}{ x} =1 $$

The question is not whether the numerator goes to $0$ (which is what you show), but whether it goes to $0$ faster than the denominator.
To see this (without L'Hopital) consider the following:
By the Taylor series, $1-\cos(x)=O(x^2)$ as $x\to 0$. 
Also, $\sin(x)=O(x)$, so $\sin(1-\cos(x))=O(x^2)$. This becomes a $O(x)$ when divided by the denominator $x$, so that the limit is $0$.
