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
 A: 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 $x_{0}$.
A: *

*What do you know about the composition of continuous functions?

*Given some $p \in \mathbb{R}$, what do you know about the function
$$
\iota_p: \mathbb{R} \longrightarrow \mathbb{R}^2,\;x \mapsto (x,p)
$$
?


These questions may seem a little mysterious/out of place to you right now, but they are actually hinting at how you are already trying to prove the statement in your question. Only, you are making implicit assumptions...
Alternatively, but very similar, let $(x_n)_n$ be a sequence in $\mathbb{R}$ with $\lim_n\, x_n = x_0$. To prove continuity of $\alpha$ at $x_0$, you have to show $\lim_n\, \alpha(x_n) = \alpha(x_0)$. Now, consider the sequence $(x_n',\,y_n')_n := (x_n,\,y_0)$ and use what you know about $f$.
Either way, you can circumvent using $\varepsilon$-$\delta$ arguments, using what you (probably) already (should) know/be allowed to use.
A: 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 
$$\lim_{x \to x_0} \alpha(x) =  \alpha(x_0)$$
