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I am searching for a methodological description on how to select the correct functions over several inequalities to find and prove that a function has a limit. I quote an example

Find the limit of the given function:

\begin{equation} \lim_{(x,y)\longrightarrow(0,0)}\frac{x^2y}{x^2+y^2} \end{equation}

The solution, in terms of inequalities , is:

\begin{equation} \bigg|\frac{x^2y}{x^2+y^2}\bigg|\leqslant|y|\leqslant\sqrt{x^2+y^2} \end{equation}

where $\sqrt{x^2+y^2}$ approaches $0$ as $(x,y)\longrightarrow(0,0)$.

First of all, where does the author of this solution get those functions between the intervals, in the solution? What rationale does he use, when he constructs these functions?

Apparently , there are many different functions that have limits that reach zero, so is this selection purely intuitive, and a subjective choice? If so, what is the objective method behind this?

Thanks

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  • $\begingroup$ Are you comfortable with polar coordinates? $\endgroup$
    – Mark S.
    Commented Aug 14, 2021 at 18:03
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    $\begingroup$ You are asking a psychological question of what the person who proposed the solution were thinking of. However, in general, to prove that $\lim f(x, y) = L$ by definition you have to start with $|f(x,y) - L| < \varepsilon$ and think what steps could lead to that starting from $\|(x,y)\| < \delta,$ that is, ideally you want to do something like $|f(x,y) - L| \leq c \|(x,y)\|$ for some constant and then set $\delta = \varepsilon/c$ (and this $c$ can be a function of the point where the limit is taken, but it is a constant nonetheless). $\endgroup$
    – William M.
    Commented Aug 14, 2021 at 20:22
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    $\begingroup$ For example, for me it would have been more natural to bound $x^2y/\sqrt{x^2+y^2} \leq y\sqrt{x^2+y^2}$ and we can take $\delta = \min(1, \varepsilon)$ so that $\sqrt{x^2+y^2} < \delta$ implies $|y| < 1$ and $\sqrt{x^2+y^2} < \varepsilon,$ thus it also implies $\left| x^2y/\sqrt{x^2+y^2} \right| \leq \left|y\sqrt{x^2+y^2} \right| < \varepsilon.$ $\endgroup$
    – William M.
    Commented Aug 14, 2021 at 20:25

1 Answer 1

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There is not an unique general way to solve this kind of limits, the way proposed is indeed a valid one among others.

First of all, assuming that the limit could be equal to zero, it is convenient to consider the absolute value for the given expression, indeed

$$|f(x)| \to 0 \implies f(x) \to 0$$

to obtain

$$\left|\frac{x^2y}{x^2+y^2}\right|=\frac{x^2|y|}{x^2+y^2}$$

then, in order to use inequalities we can observe that

$$\frac{x^2|y|}{x^2+y^2} \le \frac{x^2|y|+y^2|y|}{x^2+y^2}=|y| \to 0$$

and conclude.

The way I prefer in this case is by polar coordinates, to obtain:

$$\frac{x^2y}{x^2+y^2} = \rho \cos^2 \theta \sin \theta \to 0$$

which is simpler but, in general, needs to be used with caution. On that last issue, refer to the related:

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    $\begingroup$ The problem/subtlety with polar coordinates is that it is easy to misuse and get incorrect conclusions, as seen in this question. In this particular question, $|\rho\cos^2\theta\sin\theta|\leq \rho$, so we have a bound uniform with respect to $\theta$ which is why everything works out. Also, the parenthetical remark "$|y|\leq \sqrt{x^2+y^2}$ seems to be not necessary" isn't true. We're trying to argue $f(x,y)\to 0 $as $(x,y)\to 0$, so strictly speaking, one has to relate $|y|$ with the norm of $(x,y)$ (it's obvious in hindsight, but necessary) $\endgroup$
    – peek-a-boo
    Commented Aug 14, 2021 at 21:37
  • $\begingroup$ @peek-a-boo Thanks a lot for your comment. Yes I agree that polar coordinates need to be used with attention, I'll take a look to your example. And I also edit the parenthetical remark. Thanks $\endgroup$
    – user
    Commented Aug 14, 2021 at 21:41

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