<s>[INST] 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 $\delta=\epsilon^{\frac{1}{3}}$...
 [/INST] 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.
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