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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.

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    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. – 2017-01-26
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    Try $y=kx$ and then take derivative w.r.t. $x$. – 2017-01-26
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    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. – 2017-01-26
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    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. – 2017-01-26
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    @Did: That's exactly *Open Ball*'s approach, is it not? – 2017-01-26
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    @user21820 Modulo the typo $x^6+2y^8$ for $x^6+4y^8$, indeed it is. – 2017-01-26

2 Answers 2

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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

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

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    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.) – 2017-01-27
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    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 $0$. – 2017-01-27
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    "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. – 2017-01-27
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    If you don't mind, I'll edit your answer to fix the first problem. – 2017-01-27
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    If you don't mind, I've edited your answer to fix the first problem. I've to go soon so I'll leave you to clarify the latter issue, namely to state what I said in my above comment. Other than that, your answer is now fine. Thanks! – 2017-01-27
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    For periodic functions we don't need to check any 'endpoints' because it's periodic. For general functions we can only do the method on a compact set and must check boundary points. – 2017-01-27
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    @user21820 I cannot thank you enough for your patience and help. Just curious ... I assume that you are a math professor. If so, where and what do you teach? -Mark – 2017-01-27
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    My last edit was just to be consistent. If we don't restrict the domain then we can claim periodicity and don't have to check endpoints. If we restrict the domain to a closed interval then we shouldn't say that it's periodic and also we check endpoints. Anyway you're welcome! I am not a mathematics professor but I do research in logic and so I'm fussy about logical correctness. =) – 2017-01-27
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    Why are we calculating the extrema (i.e. when $\frac{\partial g}{\partial \theta} = 0$, yes, I know the extrema is when this equals zero but why are we finding the extrema for the limit?) Is it that if this is an extrema then this is the largest/smallest it can be and therefore taking the absolute value of the extrema will be a good enough upper bound ? – 2017-01-28
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    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. – 2017-01-28
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    thanks! I liked both answers but I'm selecting yours as the correct choice since it was consistent with the original intent of the question, namely finding the solution by use of polar coordinates. Some argue that the other answer is more "natural", I disagree. I think this is more natural but requires more work along the way while the other answer is very algebraic and very clean and short.... but requires a somewhat hidden inequality – 2017-01-28
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    @Clclstdnt You're welcome! My pleasure. The original heuristic argument was flawed. This approach was certainly not as straightforward as openball's, in which direct application of the AM-GM inequality leads to an efficient way forward. Thank you for your patience while we got this one correct. And special thanks to user21820 for the insights. -Mark – 2017-01-28
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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 $\to 0$.

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    Of course this just works for $x>0$, but you can do an argument by symmetry. – 2017-01-26
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    Well done! Better than mine. (+1) -Mark – 2017-01-26
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    @Dr.MV thanks! Admittedly, I was stuck with the polar coordinates approach, so I came up with this. +1 for your approach as well. – 2017-01-26
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    I like both of these approaches... the other answer was the method I came up with previously but wasn't able to justify it. – 2017-01-26
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    @Clclstdnt: The other answer is wrong and is an example of a **basic error in logic**. *miracle173* already tried explaining to you but you didn't get it. $(x,y) \to (0,0)$ is **NOT** equivalent to having a path towards $(0,0)$ with angle converging anywhere. For example, let $\def\nn{\mathbb{N}}$$(x_n,y_n)_{n\in\nn}$ be an enumeration of $\nn^+ \times \nn^+$. Then $(\frac1{x_n},\frac1{y_n})_{n\in\nn}$ is a sequence of points converging to $(0,0)$ but whose angle (provably) does not converge anywhere. Nor is the angle a function of distance from origin. – 2017-01-26
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    I wonder why I got a downvote. – 2017-01-26
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    @Dr.MV: You're still making a **basic logic error**. You need to prove a statement of the form "$\forall ε\ \exists r\ \forall θ\ ( \cdots )$" but you have instead proven merely "$\forall ε\ \forall θ\ \exists r\ ( \cdots )$" which is weaker. There are ways to fix your method, but ultimately they end up more complicated than *Open Ball*'s, which contains no error whatsoever (as long as one also reads his first comment). – 2017-01-26
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    @OpenBall: I also don't see a valid reason for your original answer to get a downvote, unless the reader missed your comment. – 2017-01-26
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    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 $x \le 0$. – 2017-01-26
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    @OpenBall for the record, I did not downvote. I think this is a very good answer and I understand it. – 2017-01-28
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    @Clclstdnt I'm glad you find it so. I wouldn't have thought that you might downvote, as it seems to me that the downvote came from someone who missed my first comment and didn't come back to check the edit. – 2017-01-28
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    thanks. Can you comment on the concern I posted on the other answer? My question was why we wanted to compute $g_\theta=0$ – 2017-01-28
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    @Clclstdnt I didn't go through the other answer and I'm not planning to do so at the moment. Dr. MV will surely reply to you as soon as he signs in. – 2017-01-28