# Limes of function with 2 variables [duplicate]

Find $$\lim_{x,y \rightarrow 0,0} (x^2y)/(x^4+y^4)$$ so I want to do this using polar coordinates and I get for x=rcosk; y=rsink; (x,y)->(0,0)=>r->0 $$\lim_{r \rightarrow 0} (r^3cos^2k (sink)/(r^4(cos^4k+sin^4k)))$$ = $$\lim_{r \rightarrow 0} ((cos^2k)sink)/(r(cos^4k+sin^4k)$$=something/0= ∞. But using wolframalpha I get it doesn't exist, because it depends on k. I got something / r, and I wrote it up, so it depends on r, too. What am I missing?

hint

What you called "something " could be zero.

You can simply observe that

$$f(x,x)=\frac{1}{2x}$$ and

$$\lim_{x\to 0}f(x,x)=\infty=L_1$$

On the other hand

$$\lim_{x\to 0}f(x,0)=0=L_2\ne L_1$$

So, the limit doesn't exist

• Yes I will edit "sometnihg" is $$cos^2k sink$$. And I get also ∞. But it doesn't exist, and I don't know why is that so – user779537 Oct 27 '20 at 15:27
• @user779537 something/0 could be $0/0$ if $k=n\pi$ – hamam_Abdallah Oct 27 '20 at 15:29
• -I know that but considering something = $$cos^2k sink$$ and $$cos^2k sink != 0$$ Ps. by != mean different – user779537 Oct 27 '20 at 15:30
• @user779537 when you use polar coordinates, to conclude, the result should not depend on the angle $k$. this is not the case here. – hamam_Abdallah Oct 27 '20 at 15:33
• So this is not the case here, thanks a lot. It must be mistake in textbook then – user779537 Oct 27 '20 at 15:43