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 $$\lim_{x \to 0} \dfrac{\sin x}{ x} =1 $$

By your argument, $$\lim_{x\to 0}\frac{x}{x}$$ is also $0$ because as $x$ approaches zero, both $x$ and $x$ approach $0$, so the limit is $0$.
