Use epsilon-delta definition of limits to show that $\lim_{(x,y) \to (1,2)}(3x+2y-1)=6$ and $\lim_{(x,y) \to (0,0)}(x^2+y^2)=0$

Question

Use epsilon-delta definition of limits to show that

$$\lim_{(x,y) \to (1,2)}(3x+2y-1)=6$$ $$\lim_{(x,y) \to (0,0)}(x^2+y^2)=0$$

We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{\left(x-1\right)^{2}+\left(y-2\right)^{2}}<\delta \implies \left|\left(3x+2y-1\right)-6\right|<\epsilon)))$$

Taking $\left|x-1\right|,\left|y-2\right|<\delta$ follows : $$\left|\left(3x+2y-1\right)-6\right|\le 3\left|x-2\right|+\left|2y-1\right|<3(\delta+1)+(2\delta+3)$$ $$\left|\left(3x+2y-1\right)-6\right|<5\delta +6$$

Taking $ \delta \le{(\epsilon -6)}/5$ shows the claim.

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We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{x^2+y^2}<\delta \implies \left|x^2+y^2 \right|<\epsilon)))$$ $$\left|x^2+y^2 \right|=x^2+y^2$$

Taking $ \delta \le\sqrt{\epsilon }$ shows the claim.

I want to know if my solution is true,so if you have an alternative solution that would be nice to see that,but please first check mine.

Answer 1

The latter one is correct, but the former is not. A big hint is that the choice $\delta \le (\varepsilon - 6) / 5$, which is negative(!) when $\varepsilon < 6$. Your choice of $\delta$ should always be positive, so that at least one $(x, y)$ satisfies $0 < \|(x, y) - (x_0, y_0)\| < \delta$.

Instead, consider that $$|(3x + 2y - 1) - 6| = |3(x - 1) + 2(y - 2)| \le 3|x - 1| + 2|y-2|.$$ So, if we take $\delta = \varepsilon / 5$, then \begin{align*} 0 < \sqrt{(x - 1)^2 + (y - 2)^2} < \delta &\implies |x - 1|, |y - 2| < \frac{\varepsilon}{5} \\ &\implies |(3x + 2y - 1) - 6| < 3\frac{\varepsilon}{5} + 2\frac{\varepsilon}{5} = \varepsilon. \end{align*}