Check an $\epsilon-\delta$ proof of $\lim_{ (x,y) \to (3,1)} \frac{x}{y} = 3$

Question

I am going to prove $$\lim_{ (x,y) \to (3,1)} \frac{x}{y} = 3$$ by using $\epsilon-\delta$ argument.

My Attempts

Preliminary Analysis

For every $\epsilon >0$, we need to find a suitable $\delta$ such that:

If $\Vert (x,y)-(3,1)\Vert _2 < \delta$ then $\left| \frac{x}{y} - 3\right| < \epsilon.$

First note that $$\left| \frac{x}{y} - 3\right| = \left| \frac{x}{y} - 3y + 3y -3\right| \le \left|\frac{1}{y}\right||x-3| + 3|y-1|.$$

It is obvious that $$|x-3| \le \Vert (x,y)-(3,1)\Vert_2\; \text{and} \; |y-1| \le \Vert (x,y)-(3,1)\Vert_2.$$

Suppose $\Vert (x,y)-(3,1)\Vert_2 < \frac{1}{2}$ then $|y-1| <\frac{1}{2}.$ As a result we have $$\frac{1}{2} < y < \frac{3}{2} \Longrightarrow \frac{1}{|y|} < 2.$$

Hence, $$\left| \frac{x}{y} - 3\right| \le \left|\frac{1}{y}\right||x-3| + 3|y-1|< 2\delta+3\delta = 5\delta.$$

Formal Proof

For every $\epsilon > 0$, we must choose $$\delta= \min\left\{\frac{1}{2}, \frac{\epsilon}{5}\right\}.$$

If $\Vert (x,y)-(3,1)\Vert _2 < \delta$ then $\left| \frac{x}{y} - 3\right| < \epsilon.$

Is my proof correct?

Answer 1

Your proof is not correct because

$\left|\dfrac xy-3\right|=\left|\dfrac xy-3y+3y-3\right|\leqslant\left|\dfrac1y\right||x-3|+3|y-1|$

is wrong (see my comment).

So you could proceed as follows:

$\begin{align}\left|\dfrac xy-3\right|&=\left|\dfrac 1y\right|\,\big|x-3y\big|=\left|\dfrac 1y\right|\,\big|x-3-3\big(y-1\big)\big|\leqslant\\&\leqslant \left|\dfrac1y\right|\bigg(\big|x-3\big|+3\big|y-1\big|\bigg).\end{align}$

For any $\,\varepsilon>0\,,\,$ we choose $\,\delta=\min\left\{\dfrac12,\dfrac{\varepsilon}8\right\}>0\,.$

If $\;\big\Vert(x,y)-(3,1)\big\Vert_2<\delta\,,\;$ then

$\big|x-3\big|\leqslant\big\Vert(x,y)-(3,1)\big\Vert_2<\delta\;\;,$

$\big|y-1\big|\leqslant\big\Vert(x,y)-(3,1)\big\Vert_2<\delta\;\;,$

$\left|\dfrac1y\right|=\dfrac1{\big|1+(y\!-\!1)\big|}\leqslant\dfrac1{1-\big|y\!-\!1\big|}<\dfrac1{1\!-\!\delta}\leqslant\dfrac1{1\!-\!\frac12}=2\;.$

Consequently,

$\left|\dfrac xy-3\right|\leqslant \left|\dfrac1y\right|\bigg(\big|x-3\big|+3\big|y-1\big|\bigg)<2\big(\delta+3\delta\big)=8\delta<\varepsilon\,.$