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

I am not looking for any work, just a quick yes or no answer. I have already done the work on this problem and I just want to know if I my answer is consistent with a general consensus.

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What work have you already done? – Mark Bennet Apr 5 '12 at 23:44
up vote 3 down vote accepted

If you let $x=r\cos\theta$ and $y=r\sin\theta$ (which is good for checking how the limit looks as you approach the origin from different directions), then the limit becomes $$ \lim_{r\to0}\left(r\frac{\cos^4\theta+\sin^4\theta}{\cos^3\theta+\sin^3\theta} \right)=\left\{ \matrix{ \pm\infty&\quad&\theta\to\left(k-\frac14\right)\pi\text{ for }k\in\mathbb{Z}\\\\ \text{undefined}&\quad&\theta=\left(k-\frac14\right)\pi\text{ for }k\in\mathbb{Z}\\\\ 0&\quad&\text{otherwise} } \right. $$ and since this depends on $\theta$, the limit does not exist. This is because, on the unit circle, the trigonometric ratio is well-defined except on the two points where $\cos\theta=-\sin\theta$, and at these points, the denominator vanishes while the numerator is positive, producing a ratio that blows up for all $r\ne0$ (and, in our limit, $r$ is never $0$). For other values of $\theta$, the ratio, which can also be represented as $\frac{1+\tan^4\theta}{1+\tan^3\theta}$, is a finite number, so that multiplying it by $r$ scales the result down to $0$ as $r\to0$.

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Thank you very much. This is consistent with my thinking. – shmiggens Apr 6 '12 at 0:38
Thanks. Now that you see the reasoning, if you note that $m=\tan\theta$, you might appreciate J.D.'s answer too (especially next time you have a problem like this), which is also quite nice since it gets to the point quicker. But I wanted to spell it out. – bgins Apr 6 '12 at 1:00

No. It depends on the path.

To see why, pick the path $y = mx,$ and substitute. You will get $$ x\frac{1 + m^4}{1 + m^3} $$ which depends on the path.

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But isn't the limit in this case equal to 0? – Adrián Barquero Apr 6 '12 at 0:07
@AdriánBarquero The limit is $0$ along all paths of the form: $y = mx,$ except where $m = -1.$ Which coincides with the second case in bgins' answer $\theta = (k - \pi/4).$ – user2468 Apr 6 '12 at 0:10
Oh I see, I missed that little and important detail ;) You have my upvote then =) – Adrián Barquero Apr 6 '12 at 0:12
Nice, elegant solution. You have my upvote too. – bgins Apr 6 '12 at 0:21
But does $m=-1$ give an admissible path? The function is not defined for $y=-x$. – David Mitra Apr 6 '12 at 0:34

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