How to prove that $\lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0?$ [duplicate]

Solution 1:

Note that, when $x$ and $y$ are both less than $1$, we have

$\displaystyle \left|\frac{x^2y}{x^4+y^2}\right| < \frac{1}{2}$

multiplying by $|x|$ we get

$\displaystyle \left|\frac{x^3y}{x^4+y^2}\right| < \frac{1}{2}|x|$

Result follows from squeeze theorem

Solution 2:

Cauchy inequality: $$x^4+y^2\ge 2 x^2 |y|$$

Thus $$ \left|\frac{x^3y}{x^2+y^4}\right|\le \frac{|x|}{2}. $$ But the RHS tends to $0$ as $(x,y)\to (0,0)$, and hence $$ \lim_{(x,y)\to(0,0)}\frac{x^3y}{x^2+y^4}=0. $$

Solution 3:

Observe that $x^4 + y^2 \geq |x^2y|$ (for instance, because $x^4+y^2+2x^2y = (x^2+y)^2\geq0$ and $x^4+y^2-2x^2y = (x^2-y)^2 \geq0$). Hence $\displaystyle \left|\frac{x^2y}{x^4+y^2}\right| \leq 1$ when $(x,y)\neq (0,0)$ and thus $$\lim_{(x,y)\rightarrow (0,0)} \left|\frac{x^3y}{x^4+y^2}\right| \leq \lim_{(x,y)\rightarrow(0,0)} |x| = 0,$$ so the limit is 0 by the squeeze theorem.