If $f'(x)$ exists at a point $P$, then $f(x)$ is continuous at this point. Proof.

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All the books which I've seen so far tend to proof it in a weird and over-complicated manner.

Let me try my own way.

So we know that $\lim_{\delta \to 0} \frac{f(x + \delta) - f(x)}{\delta}$ exists.

Let $f'(x) = D$.

Thus: $$\lim_{\delta \to 0}\big(f(x + \delta) - f(x)\big) = \lim_{\delta \to 0}D \times \lim_{\delta \to 0}\delta$$

Since 1) $\lim_{\delta \to 0}D = D$ and 2) infinite small sequence $\lbrace\delta\rbrace$ multiplied by constant $D$ remains infinite small:

$$\lim_{\delta \to 0}\big(f(x + \delta) - f(x)\big) = 0$$ $$\lim_{\delta \to 0}f(x + \delta) = f(x)$$ ... which is the definition of $f(x)$ being continuous at the $x$.


Question: is such a proof valid?

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As you asked, I will just point out where is not consistent to me.

  1. You set $f'(x)=D$. Thus, $D=\lim_{\delta\to 0} \frac{f(x+\delta) - f(x)}{\delta}$ is a constant. However, you used $\lim_{\delta\to 0}D$ in later proof which is not consistent with your setting.

  2. You did not say why $$ \lim_{\delta \to 0} f(x+\delta) - f(x) = \lim_{\delta\to 0}D * \lim_{\delta\to 0}\delta $$ Since this is not a conventional proof process, you may not easily get the above equality.