Is this calculus proof valid?

Question

Given is the following: A function f(x) is differentiable at some x = x_0 if the following condition holds: f(x) - f(x_0) = f'(x_0)(x-x_0)+ E(x) assuming there is an E(x) that exists such that E(x)-->0 as x--> x_0, and E(x_0) = 0. PROVE that differentiability of a single-variable function implies continuity for that function. Proof. One way of defining continuity is as follows: A function f(x) is continuous at x_0 if lim x-->x_0 f(x) exists and equals f(x_0). Taking lim x-->x_0 of both sides of the expression given in the definition of differentiability at x = x_0, we have lim x-->x_0 (f(x) - f(x_0)) = lim x-->x_0 (f'(x_0)(x-x_0)+ E(x)) Rewriting, we have lim x-->x_0 (f(x)) - f(x_0) =lim x-->x_0 f'(x)(x - x_0) + lim x-->x_0 (E(x)) By our definition of E(x) and by simple properties of limits, we simplify to lim x-->x_0 (f(x)) = 0 + lim x-->x_0 (E(x)) lim x-->x_0 (f(x)) = f(x_0) , a sufficient condition for continuity.

Answer 1

if f is differentiable in x+0 then its also continue in x_0 :

f(x) - f(x_0) = f'(x_0)(x-x_0)+ E(x)

lim x->x_0 then left side 0 and right side 0 thus f(x) = f(x_0) for x->x_0
thus continous in x_0

Answer 2

Looks right to me. A bit complicated, but correct.

Answer 3

my dear this is the defination of continuity, there is no proof. u cannot use differentiation to prove continuity. so the proof is wrong.