Limit $\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}$

Question

Find the limit

$$\displaystyle\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}$$

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Let $x=r\sin \theta$ and $y=r\cos \theta$.

$$ \begin{align} \lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}&=\lim_{r\to0}\frac{-16r^3\cos ^3 \theta r^3\sin ^3\theta}{(r^4\cos^4\theta+2r^2\sin^2\theta)^2} \\ &=\lim_{r\to0}\frac{-16r^3\cos ^3 \theta r^3\sin ^3\theta}{r^4(r^2\cos^4\theta+2\sin^2\theta)^2}\\ &=\lim_{r\to0}\frac{-16r^2\cos ^3 \theta \sin ^3\theta}{(r^2\cos^4\theta+2\sin^2\theta)^2}\\ &=\frac{-16(0)^2\cos ^3 \theta \sin ^3\theta}{((0)^2\cos^4\theta+2\sin^2\theta)^2}\\ &=\frac{0}{(0+2\sin^2\theta)^2}\\ &=0 \end{align} $$

Does this logic follow? Can I ignore the case where $2\sin^2\theta=0$ meaning it is of the form $\frac{0}{0}$? How else would I approach this limit.

Answer 1

These things get confusing, I like to work out the extreme values using Lagrange multipliers. Here, I demand $x > 0, y > 0,$ maximum value of $$ \frac{x^3 y^3}{(x^4+2y^2)^2} $$ when the denominator is thought of as the constraint. Go through the gradients being parallel, the largest value occurs when $$ 24 x^2 y^2 (x^4 + 2 y^2)(x^4 - y^2) = 0 \; . \; $$ The letters are real and positive, so $y=x^2$ gives the maximum for fixed denominator. This is then $$ \frac{x^9 }{(x^4+2x^4)^2} = \frac{x^9 }{9x^8} = \frac{x }{9} $$ That is the biggest (my) fraction can be, in the first quadrant.

I was disappointed to find that the apparent inequality suggested does not hold in general;safer to say that, when the denominator is $C>0,$ we know $x < C^{1/8}$ in the first quadrant. Then, let's see, the ratio is less than $$\frac{ C^{1/8}}{9}$$

Answer 2

By AM-GM inequality we have that

$$x^4+y^2+y^2 \geq 3 x^{\frac{4}{3}}y^{\frac{4}{3}}$$

which we can use to say

$$ \frac{16|x^3y^3|}{(x^4+2y^2)^2}\leq \frac{16|x^3y^3|}{9x^{\frac{8}{3}}y^{\frac{8}{3}}} \leq 2\sqrt[3]{|xy|} $$

Thus the limit exists and equals $0$ by squeeze theorem.

Answer 3

Another approach is to homogenize degrees with $y=ux^2$ ($u$ variable).

After substitution $x^8$ can be simplified and we get $f(x,y)=\underbrace{\dfrac{-16u^3}{(1+2u^2)^2}}_{g(u)}\times x$

It is not difficult to see that $g$ is bounded, indeed $1+2u^2\ge 1>0$ so the denominator never annulates so $g$ is continuous on $\mathbb R$.

Also $|g(u)|\sim \frac{16|u|^3}{4|u|^4}=\frac 4{|u|}\to 0$ at infinity, therefore $g$ is bounded by some $M$.

We conclude that $|f(x,y)|\le |g(u)||x|\le M|x|\to 0$

Answer 4

You already got the answer, in finding limits just avoid the intermidate $$\displaystyle\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}$$ You let $x=r\sin \theta$ and $y=r\cos \theta$.

$$ \lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2} = \lim_{r\to 0}\frac{-16r^2\cos ^3 \theta \sin ^3\theta}{(r^2\cos^4\theta+2\sin^2\theta)^2} = \frac{-16(0)^2\cos ^3 \theta \sin ^3\theta}{((0)^2\cos^4\theta+2\sin^2\theta)^2} = \frac{0}{(0+2\sin^2\theta)^2}$$

Remember that $\theta $ is also a variable $x=r\sin \theta$, so you can ignore the case when $2\sin^2 \theta$ is $0$, If fact we can deceive the equation by saying $\lim_{(\theta ,r) \to (0,0)}\frac{-16r^3\cos ^3 \theta r^3\sin ^3\theta}{(r^4\cos^4\theta+2r^2\sin^2\theta)^2}$, since $x$ and $y$ are approaching same objective $$\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2} = \lim_{x \to 0} \lim_{y \to x}\frac{-16x^3y^3}{(x^4+2y^2)^2}$$

$$\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2} = \lim_{x \to 0} \frac{-16x^3x^3}{(x^4+2x^2)^2} = \lim_{x \to 0} \frac{-16x^6}{(x^4+2x^2)^2}$$ As $x$ gets smaller the numerator shrinkes rapidly compared to the denominator, when you check L'hospitals rule after repeated sections the numerator would vanish meaning the limit is $0$