Proving $\alpha(x) = f(x,y_0)$ is a continuous function. 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

Note that
$|\alpha(x) - \alpha(x_{0})| = |f(x,y_{0}) - f(x_{0},y_{0})|$ for all suitable $x$. The function $f$ is continuous by assumption; so $|f(x,y) - f(x_{0},y_{0})|$ can be controlled on some open ball of center $(x_{0},y_{0})$. So $|f(x,y_{0}) - f(x_{0},y_{0})|$ can also be controlled on some open ball of center $x_{0}$; so $\alpha$ is continuous at $x_{0}$.
