We are supposed to prove that $\lim_{x,y \to 2,1}(3x+4y)=10$
Therefore, we need to show that $|3x+4y-10|=|3(x-2)+4(y-1)|<\epsilon$, such that for every $\epsilon$, there is some $\delta$,such that $\sqrt{(x-2)^2+(y-1)^2}<\delta.$
Now, my textbook states: "if we take $|x-2|<\delta$ and $|y-1|<\delta$, we have
$|3x+4y-10|=|3(x-2)+4(y-1)|<3|x-2|+4|y-1|<3\delta +4\delta=7\delta<\epsilon$.
So we can chose $\delta <\epsilon/7$, and the proof is finished."
Question:
My only concern is with the statement: if we take $|x-2|<\delta$ and $|y-1|<\delta$. This supposedly accounts for the fact that $\sqrt{(x-2)^2+(y-1)^2}<\delta.$ (i.e, the point $(x,y)$ is inside the circle centered at $(2,1)$ of radius $\delta$ ).
However, $|x-2|<\delta$ and $|y-1|<\delta$ only tells us that the point $(x,y)$ is within the interior of the square circumscribing the required circle. It does not necessarily imply that the point $(x,y)$ is within the circle.
How then, can we justify the fact that $\sqrt{(x-2)^2+(y-1)^2}<\delta.$?
You are confused about the definition of limit. We want $\delta$ such that $\sqrt {(x-2)^{2}+(y-1)^{2}} <\delta$ implies $|3x+4y-10| <\epsilon$. Choose $\delta$ the way the book suggests. Now start with $(x,y)$ such that $\sqrt {(x-2)^{2}+(y-1)^{2}} <\delta$. Then $|x-2| <\delta$ and $|y-1| <\delta$. Hence, $|3x+4y-10| <\epsilon$.
[I Have used the fact that $|x-2| \leq \sqrt {(x-2)^{2}+(y-1)^{2}}$ and $|y-1| \leq \sqrt {(x-2)^{2}+(y-1)^{2}}$].