Differentiable implies continuity

Question

I am slightly confused about the epsilon-delta proof of this theorem. I fully understand up to the point that: $$|f(x)-f(a)| < (|′()|+)⋅|−|$$ We then pick delta to be the $\hat{} =min(,̂/ (|′()|+))$ Thus, $|x-a|< \hat{}$. So if $$|f(x)-f(a)| < (|′()|+)⋅|−|$$and $|x-a|< \hat{}$, that means that $$|f(x)-f(a)| < (|′()|+)⋅ \hat{}$$ If $\hat{} = ̂/ (|′()|+)$, we obtain cancellations and deduce that $|f(x)-f(a)|< ̂$ for any choice not equal to initial epsilon.

If $\hat{}= $, is this the correct way to obtain continuity: $$\begin{align*}|f(x)-f(a)| &< (|′()|+)⋅|−| \\ &< (|′()|+)⋅ \\ &\leq (|′()|+)*̂/ (|′()|+) \\ &= ̂\end{align*}$$

I am using the proof on this link : https://math.stackexchange.com/questions/269666/how-to-prove-differentiability-implies-continuity-with-epsilon-delta-definit — jeff123, May 24 '20 at 01:57

score 1 · Answer 1 · answered May 24 '20 at 02:47

It doesn't matter what $\hat{\delta}$ is equal to. All we care about is an estimate: \begin{align} 0 < |x-a| < \hat{\delta} \leq \dfrac{\hat{\epsilon}}{|f'(a)| + \epsilon} \end{align}

Hence, we get the desired inequality of $|f(x) - f(a)| < \hat{\epsilon}$.

Just a small tidbit to keep in mind for the future: analysis is mostly about inequalities, not equalities. So, once you get a valid estimate, just use it.

Differentiable implies continuity

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