Limit of a function of two variables: $\lim_{(x,y) \to 0}\dfrac{x^2y}{17x^2+y^2}$

Question

$$\lim_{(x,y) \to 0} \dfrac{x^2y}{17x^2+y^2}$$

I want to obtain this limit but don't know how to. The most general advice I've found is to convert this function into polar coordinates, so when I do that I get $$\lim_{r \to 0} \dfrac{r \cos^2 x \sin x}{17\cos^2 x+ \sin^2 x}$$

but I don't think this is correct either. What should I do?

Answer 1

You are almost there, you may then observe that $$ \left|\dfrac{r \cos^2 x \sin x}{17\cos^2 x+ \sin^2 x}\right|=\left|\dfrac{r \cos^2 x \sin x}{16\cos^2 x+ 1}\right|\leq \left|r \cos^2 x \sin x\right| \leq r $$ and let $r\to 0$.

Answer 2

Hint: For any $(x,y) \neq (0,0)$, we have $0 \le \dfrac{x^2}{17x^2+y^2} \le \dfrac{1}{17}$.

Hence, $-\dfrac{1}{17}|y| \le \left|\dfrac{x^2y}{17x^2y}\right| \le \dfrac{1}{17}|y|$ for all $(x,y) \neq (0,0)$.

Now, apply the squeeze theorem.

Answer 3

Yes, it is good. The bottom is clearly $\ge 1$, and $|\cos^2 x\sin x|\lt 1$. So as $r\to 0$, the expression goes to $0$.

Answer 4

You can also use AM-GM inequality, so that $$ \left|\frac{x^2y}{17x^2+y^2}\right| \le \left|\frac{x^2y}{2\sqrt{17}xy}\right|=\frac{|y|}{2\sqrt{17}}. $$