I see that the derivative of the function is $cos(x)$ and in $[0,1]$ this can take the value $cos(0)=1$ implying that is not a contraction. Is this correct ?
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This is correct, and it is the "short version" of pre-kidney's answer. Note, indeed, that $\lim_{x\to 0}\frac{\sin x}{x}$ is another way of writing the derivative of $\sin$ at $0$. – Giuseppe Negro Nov 13 '19 at 09:28
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If $\sin x$ was a contraction on $[0,1]$ then there would exist a constant $k\in [0,1)$ such that $|\sin x-\sin y|\leq k|x-y|$ for all $x,y\in [0,1]$. To show that this is not the case, we just need a way of picking $x$ and $y$ to contradict the inequality. In fact we can choose $y=0$ and any $x$ sufficiently close to, but not equal to, $0$ such that $$\frac{\sin x}{x}>k.$$ And this, in turn, can always be done, since $$ \lim_{x\to 0}\frac{\sin x}{x}=1, $$ as is famously known.
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so i just need to pick 2 numbers that dont respect this condition and it is done, thanks – Long Claw Nov 13 '19 at 09:27
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yes, for every $k$ you would need to find $x,y$ that don't satisfy the condition. – pre-kidney Nov 13 '19 at 09:28
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3I sometimes wonder why L'Hôpital's rule is used in cases where the limit is simply the definition of the derivative ... – Martin R Nov 13 '19 at 09:30