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 argument is flawed in the sense that if both the numerator and denominator tends to 0 nothing much can be told about the limit.
Remember $\frac{0}{0}$ is indeterminate. What matters is how fast each of them goes to 0.
In the question $1-\cos x= 2 \sin^2\frac{x}{2}$ goes to 0 much faster since for very small values of x $\sin x\sim x$, or $2 \sin^2\frac{x}{2} \sim 2(\frac{x}{2})^2=\frac{x^2}{2}$
Hence $\sin (1-\cos x)\sim \sin(\frac{x^2}{2}) \sim \frac{x^2}{2}$
Obviously $\frac{x^2}{2}$ goes to 0 much faster than $x$
