Differentiability implies continuity of $f(x,y)$

I have seen proofs of this statement on the site however none use the definition of differentiability I am familiar with which is why I'm asking this question. My proof is as follows:
If $f(x,y)$ is differentiable at the point $(x_0, y_0)$ then: $$f(x,y) - f(x_0,y_0) = A(x-x_0) + B(y-y_0) + \alpha(x,y)(x-x_0) + \beta(x,y)(y-y_0)$$ where $A, B \in \mathbb{R}$ and $\alpha,\beta$ are functions such that $\lim_{(x,y)\rightarrow(x_0,y_0)}\alpha(x,y) = \lim_{(x,y)\rightarrow(x_0,y_0)} \beta(x,y) = 0$. From here we can take the limit:$$\lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)-f(x_0,y_0) =\\\lim_{(x,y)\rightarrow(x_0,y_0)} (A+\alpha(x,y))(x-x_0) + (B+\beta(x,y))(y-y_0) =\\ \lim_{(x,y)\rightarrow(x_0,y_0)} A(x-x_0) + B(y-y_0) = 0$$
Is this proof correct?