prove using the definition of a limit that $\lim_{(x,y)\to (1,2)} (x−y)^2=1$

Question

This question is for my college undergrad multivariable calculus course.

I want to prove using the definition of a limit that $\lim_{(x,y)\to (1,2)} (x−y)^2=1$.

What I have done: $f(x,y)=(x−y)^2$ and $L=1$. We must show that for any $\varepsilon>0$, we can find $\delta>0$ such that:

If $0<d((x,y),(1,2))<\delta$, then $|f(x,y)−L|=|(x−y)^2−1|<\varepsilon$.

We can choose $\delta=\varepsilon$, for if $d((x,y),(1,2))<\varepsilon$, then

$(x−1)^2+(y−2)^2<\varepsilon^2\Rightarrow x^2+y^2+1+4−2x−4y<\varepsilon^2$

I need to get this to $|(x−y)^2−1|<\varepsilon$ but I'm not able to get to this equation. What steps should I follow here?

Answer 1

If $d\bigl((x,y),(1,2)\bigr)<1$, then $|x-1|<1$ and $|y-2|<1$. Therefore$$|x-y-1|=|x-1+y-2+2|\leqslant4.$$So,\begin{align}|(x-y)^2-1|&=|x-y-1|\times|x-y+1|\\&\leqslant4|x-y+1|\\&=4|(x-1)-(y-2)|.\end{align}So, if now you take $\delta=\min\left\{\frac\varepsilon8,1\right\}$ and $(x,y)$ such that $d\bigl((x,y),(1,2)\bigr)<\delta$, then\begin{align}|(x-y)^2-1|&\leqslant4|(x-1)-(y-2)|\\&\leqslant4\bigl(|x-1|+|y-2|\bigr)\\&<4\left(\frac\varepsilon8+\frac\varepsilon8\right)\\&=\varepsilon.\end{align}

Answer 2

Alternative approach:

This response makes strong use of the result that when $0 < s,$ that

$$|r| < s \iff -s < r < s.$$

I like to keep things linear.

To prove:
$\forall \epsilon > 0, ~\exists \delta > 0,~$ such that

$$ 0 < \left| ~\sqrt{(x-1)^2 + (y-2)^2} ~\right| < \delta \implies | ~(x - y)^2 - 1 ~| < \epsilon. \tag1 $$

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My approach is to :

• Start with the assumption that $\delta$ is chosen.

• See where this leads in terms of $\displaystyle ~| ~(x - y)^2 - 1 ~| < ~$ some expression in terms of $\delta$.

• Use this to craft a relationship between $\delta$ and $\epsilon$.

Above algorithm illustrated below.

Warning
This response is long-winded. The idea is not merely to solve the given problem, but (instead) to provide the original poster with an algorithm that may be used against similar problems, many of which will be more complicated.

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First, I will automatically assume that $\delta < 1.$ This artificial constraint will be added to the relationship between $\delta$ and $\epsilon.$ Also, if the need arises, I may narrow this assumption, so that $\delta$ is assumed smaller than a specific number less than $1$.

I am (in effect) given that

$\displaystyle 0 < \left| ~\sqrt{(x-1)^2 + (y-2)^2} ~\right| < \delta.$

This implies that :

• $(x - 1)^2 < \delta^2 \implies |x - 1| < \delta.$

• $(y - 2)^2 < \delta^2 \implies |y - 2| < \delta.$

Note, that since $\delta$ is assumed less than $1$, one of the consequences of this assumption is that for any satisfying $(x,y)$, both $x$ and $y$ must be positive.

Then

• $|x - 1| < \delta \iff (1 - \delta) < x < (1 + \delta).$

• $|y - 2| < \delta \iff (2 - \delta) < y < (2 + \delta).$

Now, I want to determine a linear expression in $\delta$ such that $|(x - y)^2 - 1|$ must be less than this linear expression.

At this point, I will simplify things by narrowing the allowable value for $\delta.$ By requiring that $\delta < \dfrac{1}{2}$, I guarantee that the largest value of $x$, $(1 + \delta)$ is automatically smaller than the smallest value for $y$, $(2 - \delta).$

Therefore, $|x - y| = (y - x).$
Now, the analysis is simplified.

At this point, it is important to take a step back and examine the goal, which is:

$$| ~(x - y)^2 - 1 ~| < \epsilon \iff 1 - \epsilon < (x - y)^2 < 1 + \epsilon. \tag2 $$

I know, from the previous results that the smallest value of $|x - y|$ is produced by maximizing $x$ and minimizing $y$. Similarly, the largest value of $|x - y|$ is produced by minimizing $x$ and maximizing $y$.

Therefore,

$$(1 - 2\delta) = (2 - \delta) - (1 + \delta) < (y - x) < (2 + \delta) - (1 - \delta) = (1 + 2\delta).$$

Therefore,

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

Now, near the start of my analysis, I arbitrarily decided that $\delta < 1.$ Later, I changed this to $\delta < \dfrac{1}{2}.$ Now, I will change it again to $\delta \leq \dfrac{1}{10}.$

There are two things to notice here.

• By specifying $\delta \leq \dfrac{1}{10}$, rather than $\delta < \dfrac{1}{10}$, I made the specification compatible with the upcoming $~\displaystyle \delta = \min\left[f(\epsilon), \dfrac{1}{10}\right]~$ specification.

• Since $\delta \leq \dfrac{1}{10},~$ I know that $4 \delta^2 < \delta.$

Therefore, as a result of (3) above, I can now conclude that

$$1 - 4\delta< (y - x)^2 < 1 + 5\delta. \tag4 $$

Comparing (2) and (4) above, it is clear that a satisfying specification for $\delta$ is:

$$\delta = \min\left[\dfrac{\epsilon}{5}, \dfrac{1}{10}\right]. \tag5 $$

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In review:

The analysis in this answer establishes that when each of the following constraints are satisfied:

• $\displaystyle 0 < \left| ~\sqrt{(x-1)^2 + (y-2)^2} ~\right| < \delta$

• $\delta \leq \dfrac{1}{10}$

that

$$1 - 4\delta< (y - x)^2 < 1 + 5\delta. $$

Therefore, if I also have that $5 \delta \leq \epsilon$, I can conclude that

$$1 - \epsilon < 1 - 4\delta < (y - x)^2 < 1 + 5 \delta \leq 1 + \epsilon, $$

as required.

Answer 3

No proof is needed as $(x-y)^2$ is obviously a continuous function. The limit at $(1,2)$ is then its value at that point, which is $(-1)^2 = 1$.