Proving $\varepsilon$-$\delta$ for multivariables

Question

I have a pracitce question that I have been troubling proving: $\lim_{(x, y) \to (2, 1)} \dfrac{x^2 - 2xy}{x^2 - 4y^2} = \dfrac{1}{2}$ using the $\varepsilon$-$\delta$ definition. But here is my attempt of the solution.

If $\displaystyle \lim_{(x, y) \to (2, 1)} \dfrac{x^2 - 2xy}{x^2 - 4y^2} = \dfrac{1}{2}$, then we say that $\forall \varepsilon > 0 \exists \delta > 0 \forall(x, y) \in \mathbb{R}^2$, if $\sqrt{(x - 2)^2 + (y - 1)^2} < \delta$, then we want to show that $\displaystyle \left|\dfrac{x^2 - 2xy}{x^2 - 4y^2} - \dfrac{1}{2}\right| < \varepsilon$.

First, I tried to apply some algebraic manipulation for the given expression: \begin{equation} \left|\dfrac{x^2 - 2xy}{x^2 - 4y^2} - \dfrac{1}{2}\right| = \left|\dfrac{x}{x + 2y} - \dfrac{1}{2}\right| = \left|\dfrac{2x - (x + 2y)}{2(x + 2y)}\right| = \left|\dfrac{x - 2y}{2(x + 2y)}\right| = \dfrac{|x - 2y|}{2|x + 2y|} \end{equation} I am a little confused on how to proceed from here, so I would like some assistance. Thanks

Answer 1

$\lim_{(x, y) \to (2, 1)} \dfrac{x^2 - 2xy}{x^2 - 4y^2} = \dfrac{1}{2}$

Actually, the above assertion is false, for the following reason. In any (small) neighborhood of radius $\delta$ around $(2,1)$, you will find $(x,y) \neq (2,1)$ such that $x - 2y = 0$. If you examine the original fraction that is being interrogated:

$\displaystyle \frac{x^2 - 2xy}{x^2 - 4y^2} = \frac{x(x - 2y)}{(x + 2y)(x - 2y)}$

The assertion that the above fraction equates to $\displaystyle \frac{x}{x + 2y}$ is false, for any $(x,y) \neq (2,1)$ in the small neighborhood, where $x = 2y$.

Therefore, the only way to proceed is to add the stipulation that with

$\displaystyle f(x,y) = \frac{x^2 - 2xy}{x^2 - 4y^2}$,

you are only being asked to consider the limit as $(x,y) \to (2,1)$ for those specific values of $(x,y)$ such that $f(x,y)$ is well defined. For the present problem, this signifies the constraints of :

• $0 < |(x,y) - (2,1)| < \delta$
• $(x - 2y) \neq 0.$

Further, it is understood that although both the numerator and denominator above are going to $(0)$ as $(x,y) \to (2,1),~$ an $\varepsilon, \delta$ approach is required, rather than (for example) L'Hopital's Rule.

With the added stipulation on $f(x,y)$ discussed above, I agree with your algebraic manipulation and will use that as a starting point.

This means that if $~\displaystyle \frac {|x - 2y|}{|2(x + 2y)|} < \varepsilon$

then $~\displaystyle \left| ~\frac{x^2 - 2xy}{x^2 - 4y^2} - \frac{1}{2} ~\right| < \varepsilon$.

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$\underline{\textbf{Preliminary Results}}$

1. In general, if you have a function $f(r)$, and a positive number $A$,
   then the constraint $|f(r)| < A$ is equivalent to
   $-A < f(r) < A.$

2. The constraint that $0 < |(x,y) - (2,1)| < \delta$,
   implies that all three of the following constraints are satisfied:
   $|x - 2| < \delta$
   $|y - 1| < \delta$
   $(x,y) \neq (2,1)$.

3. As indicated, the constraint to be satisfied is
   $~\displaystyle \frac {|x - 2y|}{|2(x + 2y)|} < \varepsilon$.

4. To make things very easy on myself, I will impose the artificial constraint that
   $\delta \leq (1/10)$.

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Focusing on #2 above, you have that

• $-\delta < (x - 2) < \delta \implies (2 - \delta) < x < (2 + \delta).$
• Similarly, $(1 - \delta) < y < (1 + \delta)$.

Examining the numerator in #3 above, you have that the following lower and upper bounds (not necessarily the tightest bounds possible) are established:

$(2 - \delta) - 2(1 + \delta) < (x - 2y).$
This simplifies to $-3\delta < (x - 2y)$.

$(2 + \delta) - 2(1 - \delta) > (x - 2y).$
This simplifies to $3\delta > (x - 2y)$.

Examining the denominator in #3 above, you have that the following lower and upper bounds (not necessarily the tightest bounds possible) are established:

$2[(2 - \delta) + 2(1 - \delta)] < 2(x + 2y)$.
This simplifies to $8 - 6\delta < 2(x + 2y)$.

$2[(2 + \delta) + 2(1 + \delta)] > 2(x + 2y)$.
This simplifies to $8 + 6\delta < 2(x + 2y)$.

Putting this all together, you have that :

• $-3\delta < (x - 2y) < 3\delta$
• $8 - 6\delta < 2(x + 2y) < 8 + 6\delta$.

However, because I have imposed the artificial constraint that $\delta \leq (1/10)$, I can immediately conclude that

$7 < 2(x + 2y) < 9.$

I can express a lower bound for the fraction in #3 above by using a lower bound for the numerator and (since the numerator is negative), a lower bound for the denominator.

Therefore, I know that

$\displaystyle \frac{-3\delta}{7} < \frac{x - 2y}{2(x + 2y)}.$

Similarly, I know that

$\displaystyle \frac{3\delta}{7} > \frac{x - 2y}{2(x + 2y)}.$

This implies that whenever constraints #2 and #4 above are both satisfied, that:

$$ \frac{-3\delta}{7} < \frac{x - 2y}{2(x + 2y)} < \frac{3\delta}{7}.\tag1 $$

It is desired that (1) above should imply that constraint #3 above is satisfied. Setting $\delta = \varepsilon$ would imply that

$$ -\varepsilon = - \delta < \frac{-3\delta}{7} < \frac{x - 2y}{2(x + 2y)} < \frac{3\delta}{7} < \delta = \varepsilon. $$

Therefore, a workable final specification is :

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