i have 2 questions -

1. how does the theorem that proves differentiability implies continuity work? I understand that we're checking to see lim of x - x0 of f(x) - f(x0) = 0, but why did he multiply and divide by (x-x0)? i understand his point about when you multiply and divide by the same number the value doesn't change, but *why* did he do it? how did he know that that was what he needed to do to solve that equation?

2. how come the function f(x) = 1/x is differentiable with f'(x) = -1/x^2 even though it is not continuous?

Question

i have 2 questions - 

1. how does the theorem that proves differentiability implies continuity work? I understand that we're checking to see lim of x -> x0 of f(x) - f(x0) = 0, but why did he multiply and divide by (x-x0)? i understand his point about when you multiply and divide by the same number the value doesn't change, but *why* did he do it? how did he know that that was what he needed to do to solve that equation? 

2. how come the function f(x) = 1/x is differentiable with f'(x) = -1/x^2 even though it is not continuous?

phi · Answer

I think we say a function a function is differentiable if its derivative exists for every value in its domain.  As 1/x does not include x=0, we are ok.

phi · Answer

A link to what you are referencing in Q1 would be useful. But perhaps this explains this better https://www.khanacademy.org/math/ap-calculus-ab/product-quotient-chain-rules-ab/chain-rule-proof-ab/v/differentiability-implies-continuity