Prove limit of three variables using (ε, δ)-definition

Question

Prove limit of three variables using (ε, δ)-definition.

$$\lim_{(x, y, z)\to (0, 1, 2)} (3x+3y-z)=1$$

I have no idea how to do this with three variables.

Answer 1

You have to prove that for every $\epsilon > 0$, there exists some $\delta > 0$ such that if $$\sqrt{x^2+y^2+z^2} < \delta$$ then $|3x+3y-z-1| < \epsilon$.

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To do that, here's a hint:

• If $\sqrt{x^2+(y-1)^2+(z-2)^2} < \delta$, then $|x|<\delta$ and $|y-1|<\delta$ and $|z-2|<\delta$.
• $3x+3y-z = 3x + 3(1+(y-1)) - (z-2) -2$

Answer 2

We need to show that for every $\varepsilon > 0$, there is a value $\delta \gt 0$ such that $\left|x - 0\right| \lt \delta, \left|y-1\right| \lt \delta, \text{and } \left|z-2\right| \lt \delta$ imply that $\left|3x + 3y - z - 1 \right| \lt \varepsilon$