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Assume $f:\mathbb{R} \to \mathbb{R}$ is a differentiable function such that $\left|f '(x) \right |< M$ for all $x \in \Bbb R$. Show that $f$ is uniformly continuous.

I have no idea on how to approach this question, can someone help me?

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    $\begingroup$ What you say in the title is not what you say in the body of the question. Are you asking two separate questions? $\endgroup$ – Andrés E. Caicedo Nov 12 '13 at 15:50
  • $\begingroup$ I've reopened and reclosed to change the on hold reason to duplicate rather than unclear. $\endgroup$ – Alexander Gruber Nov 12 '13 at 22:55
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For any $x, y \in \mathbb{R}$ with $x<y$ the Mean-value theorem gives you a $z \in (x,y)$ such that $$ f(y) - f(x) = f'(z)(y-x) $$ Since $|f'(z)| < M$, it follows that $$ |f(y) - f(x)| < M|y-x| \quad\forall x,y\in \mathbb{R} $$ Now for $\epsilon > 0$, choose $\delta = \epsilon/M$, then $$ |x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon $$

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You must show that for any $\epsilon>0$, there is some $\delta>0$ such that whenever $|x-y|<\delta,$ we have $|f(x)-f(y)|<\epsilon$.

It would probably be simplest to proceed by contradiction, so assume that it is not the case, meaning there is some $\epsilon>0$ such that for any $\delta>0,$ there exist $x,y$ with $|x-y|<\delta$ such that $|f(x)-f(y)|\ge\epsilon.$ Note that such $x,y$ must be distinct (why?), and that it follows that for any $\delta>0$ there exist distinct $x,y$ such that $$\left|\frac{f(x)-f(y)}{x-y}\right|>\frac\epsilon\delta.$$ (Do you see how we can conclude this?) Choose $\delta>0$ sufficiently small so that $\frac\epsilon\delta>M,$ and apply Mean Value Theorem for a contradiction.

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