I was trying to prove that differentiability implies continuity using little-o notation. I don't know if I'm totally right, but this is what I did so far:
For differentiability we have that: $$ f(x)=f(x_{0})+f'(x_{0})(x-x_{0})+o(x-x_{0}) $$
Instead continuity states that: $$ f(x)=f(x_{0})+o(1) $$
So we can say that: $$ f'(x_{0})(x-x_{0})+o(x-x_{0})=o(1) $$
So, using little-o definition we have that: $$ \lim_{x\rightarrow x_0} \frac{f'(x_{0})(x-x_{0})+o(x-x_{0})}{1}=0 $$
Which should be true. Am I right? Is this a valid proof? :)