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I just thought I would throw this out here, hopefully it's not too boring of a question. I am stuck attempting to evaluate the following limit (these are all real numbers): $$ \lim_{\left(\chi_1, \chi_2\right) \to \left(\alpha_1, \alpha_2\right)} \frac{ \left(\left(\chi_1 + \chi_2 \right)^2 - \left(\alpha_1 + \alpha_2\right)^2\right) \hspace{1mm}-\hspace{1mm} \frac{1}{2} \left(\chi_1^2 + \chi_2^2 - \left(\alpha_1^2 + \alpha_2^2\right)\right) }{ (\chi_1 + \chi_2 - \alpha_1 - \alpha_2)^2 } $$

So, essentially, I seek to evaluate the derivative of the function

$ f : \mathbb{R}\times\mathbb{R} \to \mathbb{R} : (x_1, x_2) \mapsto (x_1+x_2)^2 - \frac{1}{2}\left(x_1^2 + x_2^2\right), $

however I don't have a lot of experience evaluating limits of functions with a domain besides $\mathbb{R}$. Does anyone have any ideas/suggestions? It's greatly appreciated.

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    $\begingroup$ Just FYI, the expression you've written is not the derivative of $f$. The derivative of $f$ should be a vector value (or, more precisely, a linear functional on $\Bbb{R}^2$), not a scalar value. $\endgroup$
    – Theo Bendit
    Jul 28, 2019 at 1:54
  • $\begingroup$ Right, I suppose that's fair. I'm still interested in in whether this limit can be evaluated or not, though, and it sort of reminds if a derivative. $\endgroup$
    – Thomas Winckelman
    Jul 28, 2019 at 2:07
  • $\begingroup$ Pardon the typo. $\endgroup$
    – Thomas Winckelman
    Jul 28, 2019 at 2:13

1 Answer 1

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Check the iterated limits first. If they do not exist, or they exist but are not equal, then you know that the double limit does not exist.

So first let $\chi_2 \to \alpha_2$ and your limit becomes $$\lim_{\chi_1 \to \alpha_1} \frac{\left((\chi_1 + \alpha_2)^2 - (\alpha_1 + \alpha_2)^2\right) - \frac 12 \left(\chi_1^2 - \alpha_1^2)\right)}{(\chi_1 - \alpha_1)^2}$$ Replacing $x = \chi_1 - \alpha_1$ and $a = \alpha_1 + \alpha_2$, this simplifies to $$\lim_{x \to 0} \frac{\left((x + a)^2 - a^2\right) - \frac 12\left((x+\alpha_1)^2 - \alpha_1^2)\right)}{x^2} \\ = \lim_{x \to 0}\frac{\left(x^2 +2ax\right) - \frac 12\left(x^2 + 2\alpha_1 x\right)}{x^2}\\ =\lim_{x \to 0}\frac{\frac 12x + (2a - \alpha_1)}x$$ which is infinite unless $\alpha_1 = -2\alpha_2$. The other iterated limit will similarly be infinite unless $\alpha_2 = -2\alpha_1$. These can both be true only when $\alpha_1 = \alpha_2 = 0$.

So your limit diverges unless $\alpha_1 = \alpha_2 = 0$, where it turns out not to converge either.

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  • $\begingroup$ Ok I like this a lot, I had never actually heard of an iterated limit and so didn't even no where to start to attack this problem, but this I can work with! This gives me an excellent starting place for where to approach problems like this in the future. One of the best answers I've received on this website. Thank you very much. $\endgroup$
    – Thomas Winckelman
    Jul 28, 2019 at 16:56
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    $\begingroup$ "iterated limit" just means taking a limit with respect to one variable first, then the other. A double limit requires coming at the limit from all directions at once. But you can reduce it to a 1-dimensional limit by choosing a curve that passes through the limit point, and taking a 1D limit along that curve. If the double limit exists, then all 1D limits along all possible curves must converge to the same value as the 2D limit. Iterated limits are curves coming in on (or tangent to) the coordinate axes. Note that the iterated limits can exist and be equal but the double limit still diverge. $\endgroup$
    – Paul Sinclair
    Jul 28, 2019 at 17:10
  • $\begingroup$ That's awesome, I also looked at the very brief wikipedia page on the subject. Makes intuitive enough sense, though I look forward to finally proving such things in my analysis sequence next year. For the mean time, though, this provides a good "quick fix." Thanks again $\endgroup$
    – Thomas Winckelman
    Jul 28, 2019 at 17:15
  • $\begingroup$ If $\alpha_1 = \alpha_2 = 0$, the multivariate limit is indeterminate. Take for one limit $(\chi_1,\chi_2) := (x,0)$ and for the other $(\chi_1,\chi_2) := (x,x/2)$. One will converge to $\frac 1 2 $, the other to $\frac {13} {18}$. $\endgroup$
    – Sudix
    Jul 28, 2019 at 18:42
  • $\begingroup$ @Sudix - you are right. I thought I had examined it thoroughly, but apparently i messed up somewhere. $\endgroup$
    – Paul Sinclair
    Jul 29, 2019 at 4:02

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