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Does the limit $$\lim_{(x,y) \to (0,0)} \left( \frac{x^2 - 2\cos(y) + 2}{y^2 - 2\cos(x) + 2} \right) $$ exist?

I think it does and it's equal to $1$, but I don't know how to prove it. I tried to use Taylor expansion of $\cos(x)$ and $\cos(y)$, but it doesn't help me to compute the limit.

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Note that, since $\cos(t)=1-\frac{t^2}{2}+O(t^4)$ as $t\to 0$, we have that $$\frac{x^2 - 2\cos(y) + 2}{y^2 - 2\cos(x) + 2}=\frac{x^2 - 2+y^2+O(y^4) + 2}{y^2 - 2+x^2+O(x^4) + 2}=\frac{r^2+O(y^4)}{r^2+O(x^4)}=\frac{1+\frac{O(y^4)}{r^2}}{1+\frac{O(x^4)}{r^2}}$$ where $r^2=x^2+y^2$. Can you take it from here?

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  • $\begingroup$ Yes, that's what I get after using Taylor expansion. Does that mean that the limit does not exist? $\endgroup$ – Jack Jun 30 '19 at 16:36
  • $\begingroup$ @Jack: So what happens if you factor $r^2$ out of numerator and denominator? $\endgroup$ – Ted Shifrin Jun 30 '19 at 16:37
  • $\begingroup$ Oh yeah, totally forgot about the common factor, now I see that it exists and it's equal to 1, thanks a lot! $\endgroup$ – Jack Jun 30 '19 at 16:39
  • $\begingroup$ @Jack Well done! $\endgroup$ – Robert Z Jun 30 '19 at 16:40

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