# Compute $\lim_{(x,y) \to (0,0)} \frac{x^2 - 2\cos(y) + 2}{y^2 - 2\cos(x) + 2}$

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.

## 1 Answer

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?

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