Double limit and iterated limit

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.

• @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? Commented Jan 11, 2021 at 6:26
• @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. Commented Jan 11, 2021 at 6:27

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.

• Thank you so much Prof. Zkutch Commented Jan 11, 2021 at 7:17
• I usually go by a different name :-). Commented Jan 11, 2021 at 7:57

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.

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