Limit of function of two variables

Question

Let $$f(x,y)=\frac{x^4y}{x^2+y^2}$$ for $(x,y)\ne (0,0)$. I want to prove that $\lim_{(x,y)\to (0,0)}\frac{x^4y}{x^2+y^2}=0$ by definition. So, I was wondering if there is any useful bound for $$\frac{x^4|y|}{x^2+y^2}$$ in order to use the definition of limit for proving this statement. Any hint will be appreciated.

Answer 1

If $x=0,y\neq 0$, then $f(0,y)=0$ and the limit is clearly zero. Else, we have $x^2+y^2 \geq x^2$, whence $(x^2+y^2)^{-1}\leq x^{-2}$, so $$ 0 \leq \left|\frac{x^4y}{x^2+y^2}\right| \leq \left|\frac{x^4y}{x^2}\right| =\left|{x^2y}\right|\to 0 $$

Answer 2

You could use that $|x|$ and $|y|$ are both less than or equal to $\sqrt{x^2+y^2}$.

Answer 3

You can also use AM-GM: $|xy|\leq \frac{x^2+y^2}{2}$

$$\left|\frac{x^4y}{x^2+y^2}\right|\leq|x|^3\frac{x^2+y^2}{2(x^2+y^2)}=\frac{|x|^3}2\stackrel{(x,y)\to (0,0)}{\longrightarrow}0$$