Prove that the limit $\lim_{(x,y)\to(0,0)} \frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x})$ exists using $\epsilon \text{ and } \delta$ definition

Question

Prove that the limit $\lim_{(x,y)\to(0,0)} \frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x})$ exists using $\epsilon \text{ and } \delta$ definition.

I have so far reduced it algebraically to the point $\frac{y^4}{\sqrt{x^2+y^2}}$ but I am not sure if I have to use polar coordinates or something to continue, kinda lost from here. I eliminated the cosine because $\cos(\frac{1}{x})\leq 1$ so

$$\frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x}) \leq \frac{y^4}{\sqrt{3x^2+y^2}}$$

And likewise I eliminated the three from $3x^2$. Do I have to use polar coordinates like $r=x^2 + y^2$???

I did some more working and got $\delta=\epsilon^{\frac{1}{3}}$...

Answer 1

Hint:$\left|\frac{y^4}{\sqrt{3x^2+y^2}}\right|\leqslant |y^3|$

Addition:

Let's take $\forall \varepsilon >0$. As we have limit $(x,y)\to (0,0)$, then we should find $\delta>0$, such, that $\sqrt{x^2+y^2}< \delta$ implies $|f|<\varepsilon$. As we have $|y|\leqslant \sqrt{x^2+y^2}< \delta$, then it gives, that $\delta^3<\varepsilon$ is enough.

Answer 2

Hint :

Observe that $y^2 \le x^2 +y^2$ so $y^4 \le (x^2 + y^2)^2$.

Also $3x^2 +y^2 \ge x^2 +y^2$ which implies that $\frac{1}{\sqrt{3x^2 + y^2}} \le \frac{1}{\sqrt{x^2+y^2}}$

Combining both we get $$\frac{y^4}{\sqrt{3x^2 + y^2}} \le \frac{(x^2+y^2)^2}{\sqrt{x^2+y^2}} = (x^2+y^2)\sqrt{(x^2+y^2)}$$

So for given $\epsilon > 0$ you can choose $\delta <\sqrt[3]{\epsilon}$ to conclude the proof.