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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?

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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

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  • $\begingroup$ 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 $\endgroup$ – user779537 Oct 27 '20 at 15:27
  • $\begingroup$ @user779537 something/0 could be $0/0$ if $ k=n\pi$ $\endgroup$ – hamam_Abdallah Oct 27 '20 at 15:29
  • $\begingroup$ -I know that but considering something = $$cos^2k sink$$ and $$cos^2k sink != 0 $$ Ps. by != mean different $\endgroup$ – user779537 Oct 27 '20 at 15:30
  • $\begingroup$ @user779537 when you use polar coordinates, to conclude, the result should not depend on the angle $ k$. this is not the case here. $\endgroup$ – hamam_Abdallah Oct 27 '20 at 15:33
  • $\begingroup$ So this is not the case here, thanks a lot. It must be mistake in textbook then $\endgroup$ – user779537 Oct 27 '20 at 15:43

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