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The answer to the question Differentiability implies continuity - A question about the proof states that , with regards to the proof given in the original question, "there is an implicit issue of existence of limits which is being swept under the rug in the presentation you have given".

Concerning the proof given in the original question: Could someone clarify what issue there is with the existence of limits, as well as how the proof in the original question is assuming continuity first?

Concerning both the proof in the original question and in the accepted answer: could someone clarify what the difference(s) is(are) between the proof in the original question, and the proof given in the answer?

To make answering the question easier, I will copy the proofs below


Proof given in the original question:

Differentiability Definition

When we say a function is differentiable at $x_0$, we mean that the limit:

$$‎f^{\prime} ‎(x) = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$ exists.

Continuity Definition

When we say a function is continuous at $x_0$, we mean that: $$\lim_{x\to x_0} f(x) - f(x_0) = 0$$

Theorem: Differentiability implies Continuity: If $f$ is a differentiable function at $x_0$, then it is continuous at $x_0$.

Proof:

Let us suppose that $f$ is differentiable at $x_0$. Then $$ \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} = ‎f^{\prime} ‎(x) $$

and hence

$$ \lim_{x\to x_0} f(x) - f(x_0) = lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot lim_{x\to x_0} (x-x_0) = 0$$

We have therefore shown that, using the definition of continuous, if the function is differentiable at $x_0$, it must also be continuous.


Proof in the accepted answer

The assumption of differentiability at $x_0$ says that the limit

$$\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$

exists as a finite number. The limit $\lim_{x \to x_0} x-x_0$ exists and is zero regardless of our assumptions. Then the product rule for limits tells us both that $\lim_{x \to x_0} f(x)-f(x_0)$ exists, and that it is the product of the two limits above, which means it must be zero. Because the product rule also tells us that the limit exists, we do not have to assume continuity first.

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In the original question, $\lim_{x \to x_0} [f(x)-f(x_0)]$ is written down directly, it seems to imply directly that we know for sure it exists.

In the answer,

First, we know that $\lim_{x \to x_0}\frac{f(x)-f(x_0)}{x-x_0}$ exists and it is equal to a finite number, $a$.

Also, we know $\lim_{x \to x_0} (x-x_0)$ exists and it is equal to zero.

and then product rule is used, and now we know $\lim_{x \to x_0}\frac{f(x)-f(x_0)}{x-x_0}\cdot \lim_{x \to x_0} (x-x_0)$ exists and it is equal to zero.

Now we know $\lim_{x \to x_0} (f(x)-f(x_0))$ exists and it is equal to zero.

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  • $\begingroup$ I see. That helps, thanks. $\endgroup$ – user106860 Sep 16 '18 at 5:29
  • $\begingroup$ I have a question - when we say, for example, what is the limit $\lim_\limits{x\to 0}\;\frac 1x$ , is it incorrect use of notation? $\endgroup$ – AnotherJohnDoe Sep 16 '18 at 5:29
  • $\begingroup$ not sure. but it should be followed by "does not exists" $\endgroup$ – Siong Thye Goh Sep 16 '18 at 5:44

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