Using $\epsilon$-$\delta$ approach, prove that $\lim_{(x,y)\rightarrow (1,3)}\frac{x}{y}=\frac{1}{3} $

Question

How to prove that $$ \lim_{(x,y)\rightarrow (1,3)}\frac{x}{y}=\frac{1}{3}$$ By $\epsilon$-$\delta$ definition.

What i did is this:

Let $\epsilon$ be grater than zero; i need to find a $\delta(\epsilon)>0$ such that $|x/y-1/3|<\epsilon$ whenever $0<\sqrt{(x-1)^2+(y-3)^2}<\delta$.

Then, rewriting $$\bigg|\frac{x}{y}-\frac{1}{3}\bigg|=\bigg|\frac{3x-y}{3y}\bigg|= \bigg|\frac{3x-y+3-3}{3y}\bigg|= \bigg|\frac{3(x-1)+(-y+3)}{3y}\bigg| $$

Now, by triangle inequality $$\bigg|\frac{3(x-1)+(-y+3)}{3y}\bigg|\leq \frac{3|x-1|+|y-3|}{3|y|}$$ since

$$|x-1|<\sqrt{(x-1)^2+(y-3)^2}<\delta \text{ and } |y-3|<\sqrt{(x-1)^2+(y-3)^2}<\delta$$

then

$$\frac{3|x-1|+|y-3|}{3|y|}<\frac{3\delta+\delta}{3|y|}=\frac{4\delta}{3|y|}$$

But i have that $|y-3|<\sqrt{(x-1)^2+(y-3)^2}<\delta$, then $y\in (3-\delta,\delta+3)$.

So i think the next step is to take $\delta =1/3$; if i do that i get $8/3<y$ that is $1/|y|<3/8$ then $$\frac{4\delta}{3|y|}<\frac{3}{8}\frac{4\delta}{3}=\frac{\delta}{2}.$$

So, taking $\delta=\min\{1/3, 2\epsilon\}$, we have

$$\bigg|\frac{x}{y}-\frac{1}{3} \bigg|< \frac{3}{8}\frac{4\delta}{3} = \frac{\delta}{2}= \frac{1}{2}2\epsilon = \epsilon. $$

is this reasoning right?

Answer 1

More generally: assume $\delta < 3$. Since you are working on an open circle of centre $(1,3)$ and radius $\delta$, you can say that $\lvert y \rvert > 3-\delta$. This implies $$\frac{4\delta}{3\lvert y \rvert} < \frac{4\delta}{3(3-\delta)},$$ and then you can set $\delta := \frac{9\varepsilon}{4+3\varepsilon}$ to find that $\left\lvert \frac{x}{y}-\frac{1}{3} \right\rvert < \varepsilon.$ Explicitly: $$\frac{4\delta}{9-3\delta} = \frac{4\frac{9\varepsilon}{4+3\varepsilon}}{9-3\frac{9\varepsilon}{4+3\varepsilon}} = \frac{36\varepsilon}{36+27\varepsilon-27\varepsilon} = \varepsilon.$$ You can now check that $$\delta = \frac{9\varepsilon}{4+3\varepsilon} < \frac{3(4+3\varepsilon)}{4+3\varepsilon} = 3$$ which shows that our choice of $\delta$ fits with the assumption $\delta < 3$.

On the other hand, if $\delta \geq 3$ you can choose points in the circle with $\lvert y\rvert$ small enough so that $\left\lvert \frac{x}{y}-\frac{1}{3} \right\rvert$ is arbitrarily large. To see this, fix $x=1$ and observe that $$\left\lvert \frac{x}{y}-\frac{1}{3} \right\rvert = \left\lvert \frac{3-y}{3y} \right\rvert \to +\infty, \text{ when } y \to 0.$$

Answer 2

You may also use this: if $$|x-1|<\epsilon\\|y-3|<\epsilon$$ we may bound $|\dfrac{x}{y}-\dfrac{1}{3}|$ sufficiently close to zero. To show that we have $$1-\epsilon<x<1+\epsilon\\3-\epsilon<y<3+\epsilon\\$$choosing $\epsilon$ small enough we obtain$$\dfrac{1-\epsilon}{3+\epsilon}<\dfrac{x}{y}<\dfrac{1+\epsilon}{3-\epsilon}$$or$$\dfrac{1}{3}-\dfrac{4\epsilon}{9+3\epsilon}=\dfrac{1-\epsilon}{3+\epsilon}<\dfrac{x}{y}<\dfrac{1+\epsilon}{3-\epsilon}=\dfrac{1}{3}+\dfrac{4\epsilon}{9-3\epsilon}$$therefore$$-\dfrac{1}{3}\epsilon<\dfrac{x}{y}-\dfrac{1}{3}<\dfrac{2}{3}\epsilon$$as long as $\epsilon<1$