Multivariable epsilon delta limit $f(x,y) = 3x + 4y $ as $(x,y) \to (1,2)$

Question

Given that $f(x,y) = 3x+4y$, I need the limit $(x,y) \to (2,1)$. I know limit is 10, so $$|3x+4y -10| = |3(x-2)+4(y-1)| < 3|x-2|+4|y-1|$$

Here we know $|x-2| < \delta = \sqrt {(x-2)^2 + (y-1)^2}$ and same for $|y-1|$, so we say inequality is satisfied when

$$7\delta \lt \epsilon$$

or when $\delta < \epsilon / 7$.

Is this correct proof?

Answer 1

$\forall \epsilon > 0$, choose $\delta = \frac{\epsilon}{7}$, if $\sqrt{(x-2)^2+(y-1)^2} < \delta, $ then $$|x-2| \color{blue}\le \sqrt{(x-2)^2+(y-1)^2} < \delta$$ and $$|y-1| \color{blue}\le \sqrt{(x-2)^2+(y-1)^2} < \delta$$

which implies that $$|3x+4y -10| = |3(x-2)+4(y-1)| \color{blue}\le 3|x-2|+4|y-1|< 7 \delta=\epsilon.$$

It is alright to choose a smaller $\delta$ too.

Answer 2

You have the correct idea about $$\delta < \epsilon / 7$$

but I would avoid statements like $$ |x-2| < \delta = \sqrt {(x-2)^2 + (y-1)^2}$$ If you rewrite your proof, it will be a very good one.