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$$ \lim_{(x,y)\to (0,0)} \frac {x^3y^2}{x^4+y^6} $$ Does this limit exist?

I've tried substituting y=x^0.5 and y=x^(2/3) which both goes to 0.

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  • $\begingroup$ Hint:Try to Change $x, y $from Cartesian to polar coordinate $\endgroup$ – user354387 Sep 19 '15 at 22:21
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    $\begingroup$ Related. The same unsatisfactory attempt to use polar coordinates was proffered as an answer there. $\endgroup$ – Jyrki Lahtonen Sep 19 '15 at 23:15
  • $\begingroup$ For a general solution : math.stackexchange.com/questions/66226/… $\endgroup$ – Arnaud D. Mar 9 '18 at 11:10
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Let $A>0$ be a parameter. Consider the curve $C_A$ determined by the equation $$ A=x^4+y^6. $$ This curve is not at all circular - it becomes quite oblong when $A\to0$. Anyway let's use these instead of circles as the polar coordinates don't seem to lead to a convincing argument.

Because both exponents are even, on the curve $C_A$ we have $|x|\le A^{1/4}$ and $|y|\le A^{1/6}$. Therefore $$ |x^3y^2|\le A^{3/4+1/3}=A\cdot A^{1/12} $$ for all points $(x,y)\in C_A$. This means that $$ |f(x,y)|=\frac{|x^3y^2|}{x^4+y^6}\le A^{1/12} $$ for all the points $(x,y)\in C_A$. This is all well, because $A^{1/12}\to0$ when $A\to0$.

The interiors of the curves $C_A$ actually form a basis of neighborhoods of the origin, so we could already conclude that the limit exists and is equal to zero. To make this more precise consider a disk $B((0,0);r)$ of radius $r>0$ around the origin. Inside that disk we have $x^4\le r^4$ and $y^6\le r^6$, so we see that $B((0,0);r)$ is contained in the union of the curves $C_A$ with $A$ ranging over the interval $0\le A\le r^4+r^6$.

This means that for a point $(x,y)\in B((0,0);r)$ we have the estimate $$ |f(x,y)|\le (r^4+r^6)^{1/12}. $$ Because $$ \lim_{r\to0+}(r^4+r^6)^{1/12}=0 $$ the sandwich principle then implies that $$ \lim_{(x,y)\to(0,0)}f(x,y)=0. $$

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  • $\begingroup$ So we got a crude but sufficient upper bound $|f(x,y)|\le C r^{1/3}$ for some constant $C$ and for all points $(x,y)\in B((0,0);r)$. Clark gets a sharper bound, constant times $r^{1/2}$, with more sophisticated estimates. $\endgroup$ – Jyrki Lahtonen Sep 20 '15 at 7:29
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We have $$ \frac{|x|^3y^2}{ x^4 + y^6 } \leq c \sqrt{ |y|}$$ Indeed it suffices to show that $$ |x|^3y^2\leq c \sqrt{ |y|} (x^4 + y^6) $$ Which we see it holds from AM - GM on $$ x^4/3 +x^4/3 +x^4/3+y^6 \geq C|x|^3\sqrt{|y|^3} $$

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    $\begingroup$ Good job! I tried to use AM-GM, but I failed to see this trick. $\endgroup$ – Jyrki Lahtonen Sep 19 '15 at 23:30

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