<s>[INST] Limit in two variables using epsilon-delta In a class assignment, I was asked to find the following limit:
$$\underset{(x,y)\to (0,0)}{\lim}\frac{1-\cos(x^2\cdot y)}{x^6+ y^4}$$
I computed it to be $0$ numerically and as such, I wanted to prove it using the $\varepsilon-\delta$ definition, but I have not been able to find a $\delta$ such that $\left\vert{\sqrt{x^2+y^2}}\right\vert < \delta$ implies $\left\vert{\frac{1-\cos(x^2\cdot y)}{x^6+ y^4}}\right\vert < \varepsilon$. I also tried using the squeeze theorem, but I have not been able to find an upper-bound for $\frac{1-\cos(x^2\cdot y)}{x^6+ y^4}$ that does not go to infinity as $(x,y)$ goes to $(0,0)$.
Any help or clue would be greatly appreciated. Thank you in advance!
 [/INST] We can use that
$$\frac{1-\cos(x^2\cdot y)}{x^6+ y^4}=\frac{1-\cos(x^2\cdot y)}{(x^2\cdot y)^2}\frac{(x^2\cdot y)^2}{x^6+ y^4}$$
with
$$\frac{1-\cos(x^2\cdot y)}{(x^2\cdot y)^2} \to \frac12$$
then for a proof by $\varepsilon-\delta$ it suffices to consider
$$\underset{(x,y)\to (0,0)}{\lim}\frac{(x^2\cdot y)^2}{x^6+ y^4}$$
and we can use that by AM-GM
$$\frac{(x^2\cdot y)^2}{x^6+ y^4}\le \frac{(x^2\cdot y)^2}{2|x|^3 y^2}=\frac{|x|}2 \to 0$$
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