How to prove that $f(x,y)=3+2x+y$ is continuous?

Question

The question is to prove that the function $f(x,y,z) = 3+2x+y$ is continuous everywhere. My approach uses the delta-epsilon method. $|(x,y)-(a,b)|\lt \delta$ then $|f(x,y)-f(a,b)|$.

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All I did was trying to follow the method, but when I got $$|3+2x-y-3-2a+b|=|(2x-2a)-(y-b)|$$ I dont know what to do after this.

Answer 1

$|f(x,y)-f(a,b)|=\\=|2(x-a)+y-b|\leq \\ \leq 2|x-a|+|y-b|\leq \\ \leq 2(|x-a|+|y-b|)\leq \\ \leq 4\sqrt{(x-a)^{2}+(y-b)^{2}}=4|(x,y)-(a,b)|<\epsilon$

if

$|(x,y)-(a,b)|<\frac{\epsilon}{4}$

Answer 2

Use $|u+v|^2 \le 2(u^2+v^2) $ with $u = x-a$ and $v = y-b$.