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The question is really simple but I'm not sure how can I prove it.

Let $f : \mathbb{R} \rightarrow \mathbb{R} \phantom{2}$ a function that verifies :

$\exists\, K \in \mathbb{R^+}, \phantom{1}\forall\, x,y \in \mathbb{R}: \lvert f(y)-f(x) \rvert \le K\lvert \cos y - \cos x \rvert$

Then f is differentiable at $0$

I proved that f is a Lipschitz function since by mean value theorem $$ \forall x,y \in \mathbb{R} \phantom{3}:\frac{|\cos y - \cos x|}{|y-x|}\le 1 \implies |\cos y - \cos x|\le|x-y| $$

Then $$\exists K \in \mathbb{R^+} \forall x,y \in \mathbb{R}: |f(y)-f(x)| \le K|\cos y - \cos x|\le K|x-y|$$

So for $x = 0$ we have that $|f(0)-f(x)| \le K|0-x|$

But I'm stilled confused on how can I apply this inequality on
$$ \lim_{x\to c} \frac{f(x) - f(c)}{x-c} $$

And if it's differentiable what's the value of $f′(0)$?

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Observe that $$ \lvert f(y)-f(x) \rvert \le K\lvert \cos y - \cos x \rvert;\forall\, x,y \in \mathbb{R},\\ \\ \implies \frac{| f(y)-f(x)|}{|y-x|}\le K\frac{\lvert \cos y - \cos x \rvert}{|y-x|}, \forall x\ne y $$ $\therefore \lim_{x\to 0} \frac{| f(0)-f(x)|}{|0-x|}\le K\cdot\lim_{x\to 0}\frac{\lvert \cos 0 - \cos x \rvert}{|0-x|}=K\cdot |\sin 0|=0$. This because $\cos $ is differentiable.

So you have $\lim_{x\to 0} \frac{| f(0)-f(x)|}{|0-x|}=0$.

You know that $\lim_{x\to c}g(x)=0 \Leftrightarrow \lim_{x\to c}|g(x)|=0 $ and hence $$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x}=0$$.

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Setting $x=0$ we get for all $y \ne 0$ $$ \left| \frac{f(y)-f(0)}{y-0}\right| \le K \left| \frac{\cos(y)-1}{y}\right| \, . $$ Now show that the limit of the right-hand side for $y \to 0$ is zero (e.g. with L'Hospital's rule, or using the definition of the derivative of the cosine).

It follows that the left-hand side also tends to zero for $y \to 0$, i.e. $f'(0) = 0$.

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