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 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 $\delta=\epsilon^{\frac{1}{3}}$...

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.
