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I have been trying to find this multivariable limit problem for some time:

$$\displaystyle\lim_{(x,y)\to (0,0)}\frac{x^5+y^5}{x^2+y^4}.$$ I got on wolfram that it does not exist, but when I substitute $x=my^2$ i get the limit $0$.

I also tried a number of other subtitutions like $ x=r^2cos^2()$ and $y= rsin()$ but I dont get a result that I can determine with it whether the limit exists or not.

Is the only way in this case is to use sequences that converge to 0? Any advice towards how to solve this limit is greatly appreciated.

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  • $\begingroup$ It's enough to consider $x,y > 0$. Then, [$x\le \epsilon$ and $y \le \epsilon$] imply [$x^5 \le \epsilon^3 x^2$ and $y^5 \le \epsilon y^4$] imply the fraction is less than $\epsilon$. $\endgroup$ – mathworker21 Mar 20 '18 at 22:26
  • $\begingroup$ @Prasun Biswas Have not you lost $x^3$? $\endgroup$ – user Mar 20 '18 at 22:40
  • $\begingroup$ @MarkViola: I just noticed user's comment and realized I left out dividing by $x^3$ in the denominator and came to a wrong conclusion. I have removed the comment. $\endgroup$ – Prasun Biswas Mar 20 '18 at 22:52
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$$ xy\ne 0\implies \left|\frac{x^5+y^5}{x^2+y^4}\right|\le \left|\frac{x^5}{x^2+y^4}\right| + \left|\frac{y^5}{x^2+y^4}\right|\le|x|^3 + |y|, $$ $$ x\ne 0, y = 0\implies \left|\frac{x^5+y^5}{x^2+y^4}\right| = |x|^3, $$ $$ x = 0, y\ne 0\implies \left|\frac{x^5+y^5}{x^2+y^4}\right| = |y|. $$ In any case: $$ \left|\frac{x^5+y^5}{x^2+y^4}\right|\le|x|^3 + |y|, $$ and the limit is $0$.

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Note that since $\cos^4\theta+\sin^4 \theta\ge m>0$ we have

$$\frac{|x^5+y^5|}{x^2+y^4}\le\frac{|x^5+y^5|}{x^4+y^4}=r\frac{|\cos^5 \theta+\sin^5 \theta|}{\cos^4\theta+\sin^4 \theta}\to 0$$

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