Proving $\lim _{(x, y) \rightarrow(3,-4)} \sqrt{x^2+y^2}=5$ using $\epsilon, \delta$

Question

Prove $$\lim _{(x, y) \rightarrow(3,-4)} \sqrt{x^2+y^2}=5$$ using $\epsilon, \delta$

My try: Given an $\epsilon >0$, we need to find $\delta>0$ such that: $$\left|\sqrt{x^{2}+y^{2}}-5\right|<\varepsilon$$ Whenever $$0<\sqrt{(x-3)^{2}+(y+4)^{2}}<\delta$$

Now we have: $$\left|\sqrt{x^{2}+y^{2}}-5\right|=\left|\sqrt{x^{2}+y^{2}}-\sqrt{3^{2}+(-4)^{2}}\right|$$ But i am stuck here?

Answer 1

HINT

Let $0 < \|(x,y) - (3,-4)\|_{2} < \delta_{\varepsilon}$.

Based on such assumption, it results that

\begin{align*} \left|\sqrt{x^{2} + y^{2}} - 5\right| & = \left|\frac{x^{2} + y^{2} - 25}{\sqrt{x^{2} + y^{2}} + 5}\right|\\\\ & \leq |x^{2} + y^{2} - 25|\\\\ & = |(x - 3)^{2} + (y + 4)^{2} + 6x - 8y - 50|\\\\ & = |(x - 3)^{2} + (y + 4)^{2} + 6(x - 3) - 8(y + 4)|\\\\ & \leq \delta^{2}_{\varepsilon} + \delta^{2}_{\varepsilon} + 6\delta_{\varepsilon} + 8\delta_{\varepsilon}\\\\ & = 2\delta^{2}_{\varepsilon} + 14\delta_{\varepsilon} := \varepsilon \end{align*}

Can you take it from here?