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$f\colon [-1,1] \rightarrow \mathbb{R}$ is differentiable. I need to show $\forall \epsilon >0,\ \exists$ a polynomial P s.t. $|f(x)-P(x)|\leq \epsilon|x|$.

I think I need to approximate the difference quotient $\frac{f(x)-f(0)}{x-0}$ with a polynomial but this isn't continuous at $0$ so I can't use Weierstrass to know that such an approximating sequence of polynomials exists.

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  • $\begingroup$ @Davide: Did you change the $\lt$ to $\leq$ on purpose in your $\TeX$ edit? $\endgroup$
    – joriki
    Aug 18 '11 at 16:39
  • $\begingroup$ @Davide it was supposed to be $\leq$ so thanks! $\endgroup$
    – user9352
    Aug 18 '11 at 16:49
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According to the definition of derivative, $\lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = f'(0)$. So if you redefine this function at 0, you do get a continuous function.

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