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Say we have the function $$f: \mathbb{R} \rightarrow \mathbb{R} ,\, with \, x \mapsto x^2$$

I understand how to prove f is differentiable using $$ f'(c) = \lim_{h \rightarrow 0} \tfrac{f(c+h) - f(c)}{h}$$ by substitution. But how would you prove differentiability using the epsilon-delta definition of limits: $$\forall \epsilon>0 \,\, \exists \delta>0 \:s.t. |x-c|< \delta \implies |\tfrac{f(x) - f(c)}{x-c} - L | < \epsilon$$ Then $$f'(c) = L$$

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Hint: $$\left|\frac{f(x)-f(c)}{x-c}-2c\right|=\left|\frac{x^2-c^2}{x-c}-2c\right|=\left|\frac{(x+c)(x-c)}{x-c}-2c\right|=\left|x-c\right|$$

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    $\begingroup$ I would write it $|x-c+2c-L|$ we don't know a priori $L$, we want to show it is $2c$. $\endgroup$ – zwim Dec 11 '17 at 17:56

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