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Multivariable limit does it exist?

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I am working with a limit problem that I get that it does not exist but W|A says it does exist and it is equal to zero...
\(\displaystyle \lim_{(x,y)->(0,0)} \frac{xy^4}{x^2+x^8}\)
well I change to polar and get after simplify
\(\displaystyle \lim_{r->0}\frac{r^3\cos(\theta)sin^4(\theta)}{\cos^2( \theta)+r^6\sin^8(\theta)}\)
which say if \(\displaystyle \theta=\frac{\pi}{2}\) we Will get \(\displaystyle \frac{0}{0}\) so it does not exist? I am wrong or?

Regards,
\(\displaystyle |\pi\rangle\)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
In higher dimensions the formal definition of a limit is much that same as in the one-dimensonal case:

We say that $f:\Bbb R^n \to \Bbb R$ has the limit:

$\displaystyle \lim_{\mathbf{x} \to \mathbf{a}} f(\mathbf{x}) = L$

if, for ANY $\epsilon > 0$, there exists a $\delta > 0$ such that:

$0 < |\mathbf{x} - \mathbf{a}| < \delta$ implies that $|f(\mathbf{x}) - L| < \epsilon$.

What this means geometrically for $\Bbb R^2$ is on any suitably small disk centered at the point $(x_0,y_0)$, the value of $f(x,y)$ should stay within $\epsilon$ of $L$.

Now for the particular function you have posted here, we can consider a disk centered at the origin (the inside of a circle).

But clearly, along the $y$-axis, $f(x,y)$ is unbounded, so no such $L$ can exist.

Wolfram|Alpha cannot always be trusted.
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Wolfram|Alpha cannot always be trusted.
[JUSTIFY]While showing that a multivariate limit exists can be hard in the general case (as it is path-dependent in the limit variable, and there are infinitely many such paths beyond one-dimensional space), it should be easy (in a computational sense) to show that a multivariate limit does not exist, though, by exhibiting two distinct paths which lead to different limits, right?

I think W|A has no concept of multivariate limits and probably just computed limits one dimension after the other, I just checked and Mathematica seems to have no built-in routine for multivariate limits either. Mathematica manages to churn out "Indeterminate", but it's probably luck more than anything else.[/JUSTIFY]
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
Hello MHB,
I am working with a limit problem that I get that it does not exist but W|A says it does exist and it is equal to zero...
\(\displaystyle \lim_{(x,y)->(0,0)} \frac{xy^4}{x^2+x^8}\)
well I change to polar and get after simplify
\(\displaystyle \lim_{r->0}\frac{r^3\cos(\theta)sin^4(\theta)}{\cos^2( \theta)+r^6\sin^8(\theta)}\)
which say if \(\displaystyle \theta=\frac{\pi}{2}\) we Will get \(\displaystyle \frac{0}{0}\) so it does not exist? I am wrong or?

Regards,
\(\displaystyle |\pi\rangle\)
The use of polar coordinates leads to the limit...

$\displaystyle \lim_{ r \rightarrow 0} \frac{r^{3}\ \cos \theta\ \sin^{4} \theta}{\cos^{2} \theta + r^{6}\ \sin^{8} \theta}\ (1)$

... and it is evident that if $\displaystyle \cos \theta=0 \implies \theta = \pm \frac{\pi}{2}$ the numerator of (1) is 0 no matter which ir r> 0 so that the limit is 0. But we can conclude that the limit is 0 only we demonstrate that, no matter wich is $\theta$, the limit is 0... what we will try to do in next post...

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
The use of polar coordinates leads to the limit...

$\displaystyle \lim_{ r \rightarrow 0} \frac{r^{3}\ \cos \theta\ \sin^{4} \theta}{\cos^{2} \theta + r^{6}\ \sin^{8} \theta}\ (1)$

... and it is evident that if $\displaystyle \cos \theta=0 \implies \theta = \pm \frac{\pi}{2}$ the numerator of (1) is 0 no matter which ir r> 0 so that the limit is 0. But we can conclude that the limit is 0 only we demonstrate that, no matter wich is $\theta$, the limit is 0... what we will try to do in next post...
Now we suppose that $\displaystyle \cos \theta = a \ne 0$... also in that case the limit is 0... but also in this case we can't conclude anything because we supposed $\theta$ to be constant but it can be $\theta= \theta (r)$ so that more investigation has to be done...


Kind regards


$\chi$ $\sigma$
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Hello MHB,
I am working with a limit problem that I get that it does not exist but W|A says it does exist and it is equal to zero...
\(\displaystyle \lim_{(x,y)->(0,0)} \frac{xy^4}{x^2+x^8}\)
well I change to polar and get after simplify
\(\displaystyle \lim_{r->0}\frac{r^3\cos(\theta)sin^4(\theta)}{\cos^2( \theta)+r^6\sin^8(\theta)}\)
which say if \(\displaystyle \theta=\frac{\pi}{2}\) we Will get \(\displaystyle \frac{0}{0}\) so it does not exist? I am wrong or?

Regards,
\(\displaystyle |\pi\rangle\)
If you approach the origin along the $x$-axis then you get the limit $0$. But if you approach the origin along the curve $y = x^{1/4}$ then the limit is \(\displaystyle \lim_{x\to0}\frac{x^2}{x^2+x^8} = \lim_{x\to0}\frac1{1+x^6} = 1\). Since you can get different values by approaching the origin in different ways, it follows that \(\displaystyle \lim_{(x,y)\to(0,0)} \frac{xy^4}{x^2+x^8}\) does not exist.
 

Petrus

Well-known member
Feb 21, 2013
739
If you approach the origin along the $x$-axis then you get the limit $0$. But if you approach the origin along the curve $y = x^{1/4}$ then the limit is \(\displaystyle \lim_{x\to0}\frac{x^2}{x^2+x^8} = \lim_{x\to0}\frac1{1+x^6} = 1\). Since you can get different values by approaching the origin in different ways, it follows that \(\displaystyle \lim_{(x,y)\to(0,0)} \frac{xy^4}{x^2+x^8}\) does not exist.
This is one of those reason I always try solve in polar.. I mean how can I know along the curve \(\displaystyle y=x^{1/4}\) I Will get a diffrent limit? I cant see that with My eyes, and try plenty curve Dont feel correct to try guess some curve and hopefully find a limit that is diffrent from the other..? May I ask how could you notice \(\displaystyle y=x^{1/4}\) Will give you diffrent answer.

Regards,
\(\displaystyle |\pi\rangle\)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
If you approach the origin along the $x$-axis then you get the limit $0$. But if you approach the origin along the curve $y = x^{1/4}$ then the limit is \(\displaystyle \lim_{x\to0}\frac{x^2}{x^2+x^8} = \lim_{x\to0}\frac1{1+x^6} = 1\). Since you can get different values by approaching the origin in different ways, it follows that \(\displaystyle \lim_{(x,y)\to(0,0)} \frac{xy^4}{x^2+x^8}\) does not exist.
This is one of those reason I always try solve in polar.. I mean how can I know along the curve \(\displaystyle y=x^{1/4}\) I Will get a diffrent limit? I cant see that with My eyes, and try plenty curve Dont feel correct to try guess some curve and hopefully find a limit that is diffrent from the other..? May I ask how could you notice \(\displaystyle y=x^{1/4}\) Will give you diffrent answer.
The danger in using only the polar method is that it is possible for all the radial limits to exist and to be equal, and yet for the limit not to exist.

That is what happens in this example. The radial limits are all zero (except for the limit along the $y$-axis, which does not exist because the function is not defined there). So to see whether there might be some different answer if you approach the origin in some other way, you need to think about a situation in which $ \dfrac{xy^4}{x^2+x^8}$ stays away from $0$ even when $x$ and $y$ are very small. For that to happen, the numerator and denominator of the fraction must stay approximately the same size, in order to make their quotient nonzero. In the denominator, if $x$ is small then $x^8$ will be much smaller than $x^2$, so the denominator is approximately $x^2$. If the numerator is to be approximately equal to the denominator then we must have $xy^4 \approx x^2$. Cancelling an $x$ on both sides leads you to look at what happens when $x\approx y^4$. That is how I came to investigate what happens along the curve $y=x^{1/4}$.