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I just watched the video titled "Differentiable Implies Continuous" and I did not understand the proof. Why is it necessary to multiply and divide by x-x0? As one needs to prove Lim x tends to x0 f(x)-(x0)=0, Can't I simply put x=x0 in that equation which gives you zero. Why is additional step required. Help me if I'm wrong.

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No. You can't always just plug in x0, as is the case with piecewise functions. You have to see how the function acts as it approaches x0 from both sides. \[\lim_{x \rightarrow x _{o}^{+}}f(x)\] must equal \[\lim_{x \rightarrow x _{o}^{-}}f(x)\] must equal \[\lim_{x \rightarrow x _{o}}f(x)\] in order for \[\lim_{x \rightarrow x _{o}}f(x)-f(x _{o})=0\] to be true.
No. I am not sure about my answer but from what I understand, you are taking the difference between a limit and a function and not between the points x and x0. In addition to that you get the differential of f at x0.
If you have a function that is differentiable at x0 and you want to know if that function is also continous at that point you must prove that\[\lim_{x \rightarrow x_{0}}f(x) -f(x_{0})=0\]is true. If you now multiply that equation by:\[\frac{ x-x_{0} }{ x-x_{0} }\]which itself is 1 and does not change the equation, you get:\[\lim_{x \rightarrow x_{0}}(f(x)-f(x_{0})\frac{ x-x_{0} }{ x-x_{0} } = 0\]Now as the premises was that function f is differentiable at point x0 as\[\lim_{x \rightarrow x_{0}}\frac{ (f(x)-f(x_{0}) }{ x-x_{0} } = f \prime (x_{0})\]we get\[\lim_{x \rightarrow x_{0}}f \prime (x_{0})(x-x_{0}) = 0\]which is obviously true. So every function that is diffrentiable at point x0 is also continous at x0 eg. diffrentable implies continous.

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The only thing to add is that even though its intuitively true that every diffrentiable function must be continous it's not valid assumption unless it is proved with mathematics. To make that proof I multiplied the equation that tells us what continous means with the expression to get to the derivate because the premise was that the function is diffrentable. So in the end we got the proof but the algebra had to be done to effect the proof.

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