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I have to find some point in the domain of $f(x,y)$ such that limit goes to zero, where: \begin{equation} f(x,y)=\frac{x^2+y^2-1}{xy} \end{equation}

For inspection, I found the point $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$. But I think that I have to prove the limit \begin{equation} \lim_{(x,y)\rightarrow \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)}\frac{x^2+y^2-1}{xy}=0 \end{equation} So for $\varepsilon>0$, there exist $\delta>0$ such that \begin{equation} \left|\frac{x^2+y^2-1}{xy}\right|<\varepsilon \hspace{0.5cm} \text{when} \hspace{0.5cm} 0<\sqrt{\left(x-\frac{\sqrt{2}}{2}\right)^2+\left(y-\frac{\sqrt{2}}{2}\right)^2}<\delta \end{equation} So this is
\begin{equation} 0<\sqrt{x^2+y^2+1-\sqrt{2}(x+y)}<\delta \end{equation} but I don't see the relation to prove this limit. Do you know some way to continue?

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    $\begingroup$ You should first identify the domain $D$ of $f$ and notice that on this domain, $f$ is continuous (because continuity is preserved by sum, product, quotient when defined). Then, the limit at a point of $D$ is equal to the value at this point and no proof is needed. $\endgroup$ Commented Sep 30, 2022 at 23:08
  • $\begingroup$ @AnneBauval Yes, the problem becomes much simpler when you focus on the continuity of the numerator and the denominator, and that at the pertinent point, the numerator is $(0)$ and the denominator is non-zero. In my judgement however, the posting's title is making the legitimate request that such considerations be ignored, in favor of a back-to-basics $\epsilon,\delta$ demonstration. See my answer. $\endgroup$ Commented Oct 1, 2022 at 1:51
  • $\begingroup$ Angie, what makes you "think that [you] have to prove the limit" and that you have no other tool at your disposal than the epsilon-delta definition? As you can see from the two answers below, it is quite heavy. Which theorems were you taught to avoid a systematic appeal to such techniques? $\endgroup$ Commented Oct 1, 2022 at 4:43
  • $\begingroup$ Dont't you remember of theorems about the limit of a sum, a product, a quotient? $\endgroup$ Commented Oct 1, 2022 at 4:48

2 Answers 2

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I always find it easier to have variables in a limit approach zero. Accordingly, in

$\begin{equation} \lim_{(x,y)\rightarrow \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)}\frac{x^2+y^2-1}{xy}=0 \end{equation}$

I would let $x=v+\frac{\sqrt{2}}{2}, y=w+\frac{\sqrt{2}}{2} $ and this becomes

$\begin{equation} \lim_{(v,w)\rightarrow \left(0, 0)\right)}\frac{(v+\frac{\sqrt{2}}{2})^2+(w+\frac{\sqrt{2}}{2})^2-1}{(v+\frac{\sqrt{2}}{2})(w+\frac{\sqrt{2}}{2})}=0 \end{equation}$

Since

$\begin{array}\\ \frac{(v+\frac{\sqrt{2}}{2})^2+(w+\frac{\sqrt{2}}{2})^2-1}{(v+\frac{\sqrt{2}}{2})(w+\frac{\sqrt{2}}{2})} &=\frac{(v^2+\sqrt{2}v+\frac12)+(w^2+\sqrt{2}v+\frac12)-1}{vw+(v+w)\frac{\sqrt{2}}{2}+\frac12}\\ &=\frac{v^2+w^2+\sqrt{2}(v+w)}{vw+(v+w)\frac{\sqrt{2}}{2}+\frac12}\\ \end{array} $

as $(v, w) \to (0, 0)$, the numerator goes to zero and the denominator is bounded away from zero (this is not too hard to prove) so the quotient goes to zero.

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  • $\begingroup$ The OP (i.e. original poster) is specifically asking for an $\epsilon,\delta$ demonstration. I interpret this request to signify that the OP is not asserting that the $\epsilon,\delta$ approach is the simplest, or that such an approach builds on continuity theorems, which is normally the preferred method. $\endgroup$ Commented Oct 1, 2022 at 2:14
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    $\begingroup$ Based on my last expression, that is straightforward. I didn't want to explicitly solve it. $\endgroup$ Commented Oct 1, 2022 at 3:05
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My standard approach to a problem like this is to strive to keep things $~\color{red}{\text{linear}}~$ between $\epsilon$ and $\delta$. For this problem there are three artificial devices that facilitate this:

  • Given two points $(x_1,y_1)$ and $(x_2,y_2)$
    let $D[(x_1,y_1),(x_2,y_2)]$ denote the (non-negative) distance between the two points.
    That is,
    $D[(x_1,y_1),(x_2,y_2)] = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}.$

    The first device is that
    $0 < D[(x_1,y_1),(x_2,y_2)] < \delta \implies $
    $0 \leq |x_1 - x_2| < \delta ~~\text{and}~~ 0 \leq |y_1 - y_2| < \delta.$

  • For the second device, suppose that $A,B,C,D$ are all positive numbers.
    Further suppose that you have two variables, that I shall call $u$ and $v$.
    Further suppose that you know that
    $A < u < B~~$ and $~~C < v < D.$

    Then you can immediately conclude that

    $\displaystyle \frac{A}{D} < \frac{u}{v} < \frac{B}{C}.$

    The idea is that you minimize $~\dfrac{u}{v}~$ by minimizing its numerator and maximizing its denominator.
    You maximize $~\dfrac{u}{v}~$ by the reverse process.

  • Given $\epsilon > 0$, when you are constructing the relationship between $\delta$ and $\epsilon$, add an additional maximum-value constraint on $\delta$.

    For example, in this problem, I will impose the constraint that $\delta$ must be $\leq \dfrac{1}{10}.$
    This implies that (for example) $\delta^2 \leq \dfrac{\delta}{10}.$
    This helps to keep things linear, between $\epsilon$ and $\delta.$


The challenge is to establish a relationship between $\epsilon$ and $\delta$ such that

$$0 < D\left[ ~(x,y),\left(~\sqrt{1/2},\sqrt{1/2}~\right) ~\right] < \delta \implies ~\left|\frac{x^2 + y^2 - 1}{xy} - 0 ~\right| < \epsilon.$$

Note
There are much easier demonstrations of the posted assertion, that in no way involve $\epsilon,\delta$. For example, it is clear that in the posted question, both the numerator and denominator are continuous functions, where the limit as $(x,y)$ approaches the pertinent point of the numerator is $(0)$ and the corresponding limit of the denominator is not zero.

Depending on what theory that you were taught, this makes it game over. However, the OP (i.e. original poster) specifically mentioned that $\epsilon$ and $\delta$ should be used. So, I will attack the problem from that viewpoint.


As discussed, since the distance between $(x,y)$ and $~\displaystyle \left(~\sqrt{1/2},\sqrt{1/2}~\right) ~$ is between $(0)$ and $(\delta),$ exclusive, you have that

  • $\displaystyle 0 \leq |x - \sqrt{1/2}| < \delta < \frac{1}{10}.$

  • $\displaystyle 0 \leq |y - \sqrt{1/2}| < \delta < \frac{1}{10}.$

This implies that $x,y$ are both bounded by:

  • $\displaystyle \sqrt{1/2} - \delta < x < \sqrt{1/2} + \delta.$

  • $\displaystyle \sqrt{1/2} - \delta < y < \sqrt{1/2} + \delta.$

This implies that

$$~\frac{1}{2} - 2\delta\sqrt{1/2} + \delta^2 < x^2 < \frac{1}{2} + 2\delta\sqrt{1/2} + \delta^2.$$

Since $\delta \leq \dfrac{1}{10}$,
this implies that

$$~\frac{1}{2} - 2\delta\sqrt{1/2} < x^2 < \frac{1}{2} + \delta\left[2\sqrt{1/2} + \frac{1}{10}\right].$$

Further, since $~\displaystyle 2\sqrt{1/2} < 1.5~$
I can immediately conclude that

$$\frac{1}{2} - 2\delta < x^2 < \frac{1}{2} + 2\delta. \tag1 $$

By parallel analysis applied to the variable $(y)$, I can similarly conclude that

$$\frac{1}{2} - 2\delta < y^2 < \frac{1}{2} + 2\delta. \tag2 $$

This means that I know that the numerator,
$(x^2 + y^2 - 1)$ is bounded by

$$-4\delta < (x^2 + y^2 - 1) < 4\delta. \tag3 $$


Now, it is time to establish bounds on the denominator, $(xy)$.

Note that in the (small) neighborhood of radius $\delta$ around $~\displaystyle \left(\sqrt{1/2},\sqrt{1/2}\right),~$ that both of the coordinates of $(x,y)$ are positive.

Further, using (1) and (2) above, I have that

$$\left[\frac{1}{2} - 2\delta\right]^2 < (xy)^2 < \left[\frac{1}{2} + 2\delta\right]^2.$$

This implies [since $x,y$ are each known to be positive], that

$$\frac{1}{2} - 2\delta < (xy) < \frac{1}{2} + 2\delta. \tag4 $$

It only remains to

  • put (3) and (4) above together

  • use this to determine upper and lower bounds on the fraction, in terms of $\delta$

  • use these bounds to construct the appropriate relationship between $\epsilon$ and $\delta.$

Since $0 < \delta \leq \dfrac{1}{10}$, I can immediately conclude that

$$0.3 \leq \frac{1}{2} - 2\delta < (xy) < \frac{1}{2} + 2\delta \leq 0.7. \tag5 $$

Since $\dfrac{1}{4} < 0.3$, I can therefore, immediately conclude that

$$\dfrac{1}{4} < (xy) < 1.$$

Now, I can use this in conjunction with (3) above.

I can now conclude that

$$\frac{-4\delta}{(1/4)} < \frac{x^2 + y^2 - 1}{xy} < \frac{4\delta}{(1/4)} \implies $$

$$-16\delta < \frac{x^2 + y^2 - 1}{xy} < 16\delta.$$

Therefore, I can conclude that the following relationship between $\delta$ and $\epsilon$ will be satisfactory:

$$\delta = \min\left[ ~\frac{\epsilon}{16}, ~\frac{1}{10} ~\right].$$

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