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 $$f'(c) = L$$


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

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.