If $f$ is differntiable at $x_0$, then $f$ is continuous at $x_0$. How does this work? [duplicate]

There's a similar post here. but I haven't got the answer:
If $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.

I just started learning calculus for physics, so I don't understand well yet. I was watching an MIT OpencourseWare Lecture (link below), and I just couldn't understand the last section of the video where the proof of a theorem was being explained. I did search online for answers but didn't understand their explanations, and it feels like I'm missing something really obvious.

The theorem states: If $f$ is differentiable at $x_0$, then $f$ is continuous. The proof goes like this:

1) $$\lim_{x\to x_0} f(x)-f(x_0) = 0$$ 2) $$\lim_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}\cdot(x-x_0)=f'(x_0)\cdot0=0$$

What did I understand is that the first equation is just the definition of continuity: $\lim\limits_{x\to x_0} f(x) = f(x_0)$ rearranged. But I do not understand the second part at all.

Can someone please explain what is happening in the second equation? Also:
1) Where does the $f'(x_0)$ suddenly come from?
2) Why is it being multiplied with $\dfrac{x-x_0}{x-x_0}$?

It does look vaguely familar to the difference quotient: $\lim\limits_{\Delta x\to 0} f'(x_0) = \dfrac{f(x_0+\Delta x) - f(x_0)}{\Delta x}$

I am suspecting that it has something to do with $\Delta x$ being implied somewhere.

Video (starts at 46:10)

marked as duplicate by Roland, Masacroso, Juniven, Zain Patel, user223391 Apr 24 '17 at 4:03

• By the definition of derivative: $f'(x_0)=\lim_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ – Itay4 Apr 22 '17 at 13:45
• The second part is equivalent to the first one because it is the same but dividing and multiplying by $(x-x_0)$ what is valid because is not zero. The order is the inverse: first we show 2), then you can see that 1) is a consequence of 2). – Masacroso Apr 22 '17 at 13:45
• @Italy4 Thanks, but I don't get how the numerator, $f(x_0 + \Delta x) - f(x_0)$ becomes $f(x)-f(x_0)$ – WeavingBird1917 Apr 22 '17 at 13:48
• @WeavingBird1917 because $\Delta x$ is $(x-x_0)$ – Giulio Apr 22 '17 at 13:50
• To those who have voted to close this: I'd have to argue against closing this one. The OP is clearly new to Calculus (the other one has a real analysis tag) and it's clear that the definition of the derivative that the OP is provided is different than the one that was mentioned in the other question. – Clarinetist Apr 22 '17 at 13:54

Your definition is almost correct: it should be $$f^{\prime}(x_0) = \lim_{\Delta x \to 0}\dfrac{f(x_0+\Delta x) - f(x_0)}{\Delta x}\text{.}$$ To see how this definition can be written differently, perform a change of variables, where $x = x_0 + \Delta x$. Then the limit is taken as $\Delta x = x - x_0 \to 0$ or equivalently $x \to x_0$, hence $$f^{\prime}(x_0) = \lim_{x - x_0 \to 0}\dfrac{f(x) - f(x_0)}{x - x_0} = \lim_{x \to x_0} \dfrac{f(x) - f(x_0)}{x - x_0}\text{.}$$ Now as $x \to x_0$, $x - x_0 \to 0$. Hence $$\lim_{x \to x_0} \left[\dfrac{f(x) - f(x_0)}{x - x_0}\cdot (x-x_0)\right] = \lim_{x \to x_0} \left[\dfrac{f(x) - f(x_0)}{x - x_0}\right]\cdot \lim_{x \to x_0}(x-x_0) = f^{\prime}(x_0) \cdot 0\text{.}$$
$$\lim_{x\to x_0} (f(x)-f(x_0)) = \lim_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}\cdot(x-x_0)=f'(x_0)\cdot0=0$$ where we have used the fact that the limit of a product is the product of the limits, if both exist. Also we have substituted $x-x_0$ for your $\Delta x$ in the definition of the derivative. From this we have $$\lim_{x\to x_0} (f(x)-f(x_0)) = 0$$ which is the definition of continuity at $x_0$.