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Prove that if $f$ is differentiable at $x_{0}$ then there exists a $ \delta>0 $ and a $K_{0}>0$ such that for all $x\in N_{\delta}(x_{0})$, $|f(x)-f(x_{0})|\leq K_{0}|x-x_{0}|$

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closed as off-topic by Normal Human, Sami Ben Romdhane, martini, user1729, Mathmo123 Jul 30 '14 at 10:00

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2 Answers 2

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$$\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)\Longrightarrow \,\forall\,\epsilon >0\,\,\exists\,\delta>0\,\,\,s.t.$$

$$|x-x_0|<\delta\Longrightarrow\left|\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)\right|<\epsilon\Longleftrightarrow $$

$$|f(x)-f(x_0)|<\left(f'(x_0)+\epsilon\right)|x-x_0|\,\,,\,\,\forall x\in N_\delta(x_0):=\{x\in\Bbb R\;;\;|x-x_0|<\delta\}$$

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Note the homework tag. I like this answer on how to handle homework questions. I prefer to leave at most a hint. –  robjohn Dec 4 '12 at 18:38

Hint $$ \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0) $$ Consider the definition of a limit.

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