Find the limit of the function $f(x,y)=\frac{2x^2y}{x^4+y^2}$ at the origin.

Question

I did substitute $y=mx$. $$\lim_{(x,y)\rightarrow (0,0)} f(x, y)=\lim_{ x\rightarrow 0} \frac{2mx^3}{x^4+mx^2}\\ =\lim_{ x\rightarrow 0} \frac{2mx}{x^2+m}$$ The function $\frac{2mx}{x^2+m}$ is contineous at $x=0$, the limit is $0$.

But if I use, online calculators, it says the limit does not exists. Whats the mistake in my way?

See for example https://math.stackexchange.com/q/202604/42969 or https://math.stackexchange.com/q/93886/42969. — Martin R, Apr 17 '21 at 18:38
This https://math.stackexchange.com/q/829201/42969 demonstrates the flaw in your argument. — Martin R, Apr 17 '21 at 18:40
Does this answer your question? Continuity conditions for multivariate functions. — Martin R, Apr 17 '21 at 18:47

José Carlos Santos · Accepted Answer · 2021-04-17T18:38:08.587

4

Note that $f(x,x^2)=1$ for each $x\ne0$. The fact that $\lim_{x\to0}f(x,mx)=0$ for each $m\in\Bbb R$ is no enough to prove that the limit is $0$.

edited Apr 17 '21 at 18:38

answered Apr 17 '21 at 18:36

José Carlos Santos

427,504

So the substitution method won't work here? – Madhan Kumar Apr 17 '21 at 18:37
I have expanded my answer. But what I did is a substitution: $y=x^2$. – José Carlos Santos Apr 17 '21 at 18:38

Find the limit of the function $f(x,y)=\frac{2x^2y}{x^4+y^2}$ at the origin.

1 Answers1