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I was hanging on the following statement.

"If the double limit $\displaystyle \lim_{(x_1, x_2) \rightarrow (a_1, a_2)} f(x_1, x_2)$ and the iterated limit $\lim_{x_1 \rightarrow a_1} \lim_{x_2 \rightarrow a_2} f(x_1, x_2)$ exist, then they will be equal."

Towards this statement I was thinking the function $x \sin \tfrac{1}{y} + y \sin \tfrac{1}{x}$. But for this function, the iterated limit does not exist.

Can you give a counterexample of the statement.

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  • $\begingroup$ @Kavi: I hope no. Take $\lim_{x \rightarrow 0} (x \sin \tfrac{1}{y} + y \sin \tfrac{1}{x})$ and $\lim_{(x, y) \rightarrow (0, 0)} (x \sin \tfrac{1}{y} + y \sin \tfrac{1}{x})$. Do you have any authentic source for your statement? $\endgroup$
    – Summation
    Commented Jan 11, 2021 at 6:26
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    $\begingroup$ @Kavi Rama Murthy you are wrong. Taking $f(x,y)=x\sin \frac{1}{y}+y\sin \frac{1}{x}$ for $xy \ne 0$ and $f=0$ for $xy = 0$ gives counterexample. $\endgroup$
    – zkutch
    Commented Jan 11, 2021 at 6:27

2 Answers 2

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Suppose $\lim_{x \to \alpha} f(x) = L$ then for any $\epsilon>0$ there is some $\delta>0$ such that if $x$ satisfies $0< \|x-\alpha\| < \delta$ then $|f(x)-L| < \epsilon$.

Suppose $x_1$ satisfies $0 < |x_1 - \alpha_1 | < {1 \over \sqrt{2}} \delta$, then for $0 < |x_2 - \alpha_2 | < {1 \over \sqrt{2}} \delta$ we have $0< \|x-\alpha\| < \delta$ (with $x=(x_1,x_2)$) and so $|f(x)-L| < \epsilon$.

In particular, the two limits are equal.

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  • $\begingroup$ Thank you so much Prof. Zkutch $\endgroup$
    – Summation
    Commented Jan 11, 2021 at 7:17
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    $\begingroup$ I usually go by a different name :-). $\endgroup$
    – copper.hat
    Commented Jan 11, 2021 at 7:57
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As it is in John M.H. Olmsted - Advanced calculus-Prentice Hall (1961), page 184, the existence of double limit and either of the two iterated limits, finite or infinite, implies the equality of double and that iterated.

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  • $\begingroup$ Thank you so much Prof. Zkutch $\endgroup$
    – Summation
    Commented Jan 11, 2021 at 7:16
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    $\begingroup$ You are welcome. Feel free to ask more when needed. $\endgroup$
    – zkutch
    Commented Jan 11, 2021 at 9:25

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