TITLE: Proving $\alpha(x) = f(x,y_0)$ is a continuous function.
QUESTION [6 upvotes]: Problem: $f:\mathbb{R}^2\to \mathbb{R}$, if f is continuous at $(x_0,y_0)$ show that the function $\alpha(x) = f(x,y_0)$ is continuous at $x_0$.
I know that a function is continuous if $$\lim_{(x,y) \to (x_0, y_0)}  f(x,y) = f(x_0,y_0)$$
I did this:
$$\lim_{x \to x_0} \ \alpha(x) = \lim_{x \to x_0} \ f(x,y_0) = f(x_0,y_0) = \alpha(x_0)$$
I doubt this is okay, need help, thanks

REPLY [2 votes]: Let $\alpha(x) =  f(x,y_0)$ from continuity of $f$ we have $\forall\varepsilon>0\,\exists\delta>0$ such that if $(x-x_0)^2+(y-y_0)^2<\delta^2$ then $|f(x,y)-f(x_0,y_0)|<\varepsilon$. 
If $|x-x_0|<\delta$ then $(x-x_0)^2+(y_0-y_0)^2<\delta^2$ an thus $|f(x,y_0)-f(x_0,y_0)|<\varepsilon$ or $|\alpha(x)-\alpha(x_0)|<\varepsilon$. This means 
$$\lim_{x \to x_0} \alpha(x) =  \alpha(x_0)$$