Limit in two variables using epsilon-delta

Question

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!

Answer 1

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

$$\frac{(x^2\cdot y)^2}{x^6+ y^4}\le \frac{(x^2\cdot y)^2}{2|x|^3 y^2}=\frac{|x|}2 \to 0$$

Answer 2

We know that:

1. $\ \ \ \forall u \in \mathbb R \ ,\ 0\leqslant 1-\cos u \leqslant \dfrac{u^2}{2}\ \ $
2. $\ \ \forall (x,y) \in \mathbb R^2 \ ,\ |x^3y^2| \leqslant \dfrac{1}{2} \left( x^6 +y^4\right) $

So, we can write:

$0 \leqslant \dfrac{1-\cos\left( x^2y\right)}{x^6+y^4} \leqslant \dfrac{1}{2} \dfrac{x^4y^2}{x^6+y^4}=\dfrac{1}{2} \dfrac{\left|x^3y\right|}{x^6+y^4}|x| \leqslant \dfrac{1}{4} |x| $

And it's easy to conclude.