# Problem in showing $\lim_{(x,y)\to (0,0)} \frac{12x^3y^5+4x^4y^4}{x^6+4y^8}=0$ using polar coordinates

I'm trying to show that $$\lim_{(x,y)\to (0,0)} \frac{12x^3y^5+4x^4y^4}{x^6+4y^8}=0$$

I've used polar coordinates but when I do this I get the possibility of $\frac{0}{0}$ if $\cos(\theta)\to 0$ as $r\to 0$. So it must be that I need some sort of bound to make sense of this limit. But I'm not sure how to proceed.

• The limit is for what happens as (x,y) approaches (0,0), not (x,y) actually getting to (0,0). // Example - As x approaches zero 1/x goes to infinity, but 1/0 is undefined of course.
– MaxW
Jan 26, 2017 at 6:34
• Try $y=kx$ and then take derivative w.r.t. $x$. Jan 26, 2017 at 7:17
• What?! I think you are confused. There is no need to take a derivative. You should try to understand the polar coordinate answer below because it is very well thought out and is absolutely correct and contains in it every possible curve, including your curves $y=kx$ as special cases. Jan 26, 2017 at 7:24
• Polar coordinates are a strange idea in this context. Note that $$4|x^3|y^4\leqslant x^6+4y^8$$ hence $$\left|\frac{12x^3y^5+4x^4y^4}{x^6+4y^8}\right|\leqslant3|y|+|x|$$ and the limit follows.
– Did
Jan 26, 2017 at 17:19
• @Did: That's exactly Open Ball's approach, is it not? Jan 26, 2017 at 17:22

Hint: $u^2 + 4v^2 = (u-2v)^2 + 4uv \ge 4uv$, so:

$$x^6 + 4y^8 = (x^3)^2 + 4(y^4)^2 \ge 4x^3 y^4$$

Edit:

I pointed out in a comment that an argument by symmetry works.

A slightly better approach might be as follows:

$$\left| \frac{12x^3 y^5 + 4 x^4y^4}{x^6 + 4y^8} \right| \le 4\frac{3|x|^3 |y|^5 + x^4y^4}{x^6 + 4y^8}$$

Now we use $x^6 + 4y^8 \ge 4|x|^3 y^4$ to see that the RHS $\to 0$.

• Of course this just works for $x>0$, but you can do an argument by symmetry.
– user384138
Jan 26, 2017 at 6:42
• Well done! Better than mine. (+1) -Mark Jan 26, 2017 at 6:44
• @Dr.MV thanks! Admittedly, I was stuck with the polar coordinates approach, so I came up with this. +1 for your approach as well.
– user384138
Jan 26, 2017 at 6:47
• @OpenBall: I also don't see a valid reason for your original answer to get a downvote, unless the reader missed your comment. Jan 26, 2017 at 16:02
• One possible thing to note is that it's better to not use symmetry on the outermost level since a path towards the origin may oscillate back and forth between the two halves or on the $y$-axis. This issue can be resolved easily unlike in the other 'answer', but your method in your edit is definitely superior, which is equivalent to using symmetry on the inner level, which is what I originally thought you did on reading your comment, but perhaps you didn't. Specifically, you have $x^6 + 4y^8 \ge 4x^3 y^4$ for $x \ge 0$. By reflecting $x$ you get $x^6 + 4y^8 \ge 4(-x)^3 y^4$ for $x \le 0$. Jan 26, 2017 at 16:13

If we wish to use polar coordinates $(r,\theta)$, then we can write

\begin{align} \frac{12x^3y^5+4x^4y^4}{x^6+4y^8}&=r^2\left(\frac{12\cos^3(\theta)\sin^5(\theta)+4\cos^4(\theta)\sin^4(\theta)}{\cos^6(\theta)+4r^2\sin^8(\theta)}\right)\\\\ &=r\left(3\sin(\theta)+\cos(\theta)\right)\left(\frac{4r\cos^3(\theta)\sin^4(\theta)}{\cos^6(\theta)+4r^2\sin^8(\theta)}\right)\tag1 \end{align}

Let $g(r,\theta)=\frac{4r\cos^3(\theta)\sin^4(\theta)}{\cos^6(\theta)+4r^2\sin^8(\theta)}$. Denote $\sin(\theta)$ by $s$ and $\cos(\theta)$ by $c$.

We will view $g(r,\theta)$ as a function of $\theta\in \mathbb{R}$, which is differentiable and $2\pi$-periodic. Therefore the extrema occur at points for which $\frac{\partial g(r,\theta)}{\partial \theta}=0$.

Then, taking the partial derivative with respect to $\theta$, and , we have

\begin{align} \frac{\partial g(r,\theta)}{\partial \theta}&=4rs^3c^2\,\left(\frac{(4c^2-3s^2)(c^6+4r^2s^8)-s^2c^2(32r^2s^6-6c^4)}{(c^6+4r^2s^8)^2}\right)\\\\ &= 4rs^3c^2\,\left(\frac{(3+c^2)(c^6-4r^2s^8)}{(c^6+4r^2s^8)^2}\right)\tag 2 \end{align}

We see that $\frac{\partial g(r,\theta)}{\partial \theta}=0$ when $\sin(\theta)=0$ or $\cos(\theta)=0$ or $\cos^6(\theta)=4r^2\sin^8(\theta)$.

When $\sin(\theta)=0$ or $\cos(\theta)=0$, $g(r,\theta)=0$. When $\cos(\theta)^6=4r^2\sin^8(\theta)$,

$$g(r,\theta)=\text{sgn}(\cos(\theta))\tag 3$$

Finally, using $(3)$ in $(1)$ reveals

$$\left|\frac{12x^3y^5+4x^4y^4}{x^6+4y^8}\right|\le r|3\sin(\theta)+\cos(\theta)|$$

whereupon applying the squeeze theorem yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{(x,y)\to (0,0)}\frac{12x^3y^5+4x^4y^4}{x^6+4y^8}=0}$$

• Sorry but it is still not incorrect; though the logical error is more subtle this time. The extrema of a differentiable function on a compact set is either at a stationary point or at a boundary point. You did not specify a compact set so the method is simply wrong. You cannot even look at the function $( r \mapsto g(r,θ) )$ because it's difficult to work with at $0$. Instead it is possible to find the extrema of $( θ \mapsto g(r,θ) )$ on the compact set $[0,2π]$. (I didn't get notified of your comment; it seems pinging only works for those already in the comment thread.) Jan 27, 2017 at 2:35
• In detail, after finding the extrema of $( θ \mapsto g(r,θ) )$ on $[0,2π]$, we then know the extrema of $f$ on each circle around $0$ in terms of $r$, and then we can take limit as $r \to 0$ to find squeeze bounds on $f$ near $0$. Jan 27, 2017 at 2:45
• "We see that $g′(r,θ)=0$ when ..., ..., and ..." Nope; "or" not "and". Also, I would like you to make it very clear in your answer that $( θ \mapsto g(r,θ) )$ is differentiable and periodic on the real line, and hence the extrema must occur when its derivative is zero. If you don't, students might make the same mistake as you; it's a common mistake due to sloppy teaching of the method. Jan 27, 2017 at 15:17
• If you don't mind, I'll edit your answer to fix the first problem. Jan 27, 2017 at 15:21
• Good questions. You're spot on. At the extreme values, $g$ is either a minimum ($-1$) or a maximum ($+1$). And we can obtain, therefore, a bound on the absolute value of $f$. Then application of the squeeze theorem finishes it. Jan 28, 2017 at 20:11