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Multivariable limit

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,

\(\displaystyle \lim_{(x,y)->(0,0)} \frac{6x^3y}{2x^4+x^4}\)
I did easy solve that the limit do not exist by \(\displaystyle (0,t)=0\), \(\displaystyle (t,0)=0\), \(\displaystyle (t,t)=\frac{6}{3}\)
but I wanted Also to solve this by polar cordinate so we got
\(\displaystyle \lim_{r->0}\frac{6\cos(\theta)\sin(\theta)}{2\cos(\theta)+ \sin(\theta)}\)
so My question is what can I say to show this limit Will never exist.
My argument: the top Will never be same for \(\displaystyle \theta\) and bottom Will never be equal to zero. So the limit does not exist and this does not sound like a argument for me..

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

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
Hello MHB,

\(\displaystyle \lim_{(x,y)->(0,0)} \frac{6x^3y}{2x^4+x^4}\)
I did easy solve that the limit do not exist by \(\displaystyle (0,t)=0\), \(\displaystyle (t,0)=0\), \(\displaystyle (t,t)=\frac{6}{3}\)
but I wanted Also to solve this by polar cordinate so we got
\(\displaystyle \lim_{r->0}\frac{6\cos(\theta)\sin(\theta)}{2\cos(\theta)+ \sin(\theta)}\)
so My question is what can I say to show this limit Will never exist.
My argument: the top Will never be same for \(\displaystyle \theta\) and bottom Will never be equal to zero. So the limit does not exist and this does not sound like a argument for me..

Regards,
\(\displaystyle |\pi\rangle\)
Perhaps this will help: As r approaches 0, theta will become undefined.

-Dan
 

Petrus

Well-known member
Feb 21, 2013
739
Perhaps this will help: As r approaches 0, theta will become undefined.

-Dan
Ofcourse that's true.. I see I Was only confusing myself.. I only need to argument for like \(\displaystyle \lim_{r->0}\frac{r\cos(\theta)}{\cos(\theta)+\sin(\theta)}\) that \(\displaystyle \sin(\theta)+cos(\theta) \neq 0\) while this one is OBVIOUS that it Will not exist!
thanks For the help!

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

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Perhaps this will help: As r approaches 0, theta will become undefined.

-Dan
Not really. As Petrus found, what we have is that

$$\lim_{r \to 0^+} \frac{6 \cos^3 \theta \sin \theta}{2 \cos^4 + \sin^4 \theta} = \frac{6 \cos^3 \theta \sin \theta}{2 \cos^4 + \sin^4 \theta},$$

since it is no longer dependent on $r$. The point is: this is now completely dependent on $\theta$. Therefore, taking $\theta = 0$ we get the zero answer while taking $\theta = \pi/4$ we get another nonzero answer. Hence the limit does not exist.

Ofcourse that's true.. I see I Was only confusing myself.. I only need to argument for like \(\displaystyle \lim_{r->0}\frac{r\cos(\theta)}{\cos(\theta)+\sin(\theta)}\) that \(\displaystyle \sin(\theta)+cos(\theta) \neq 0\) while this one is OBVIOUS that it Will not exist!
thanks For the help!

Regards,
\(\displaystyle |\pi\rangle\)
I do not follow your reasoning. Even though it will not exist, what does it have to do with the fact that $\sin \theta + \cos \theta \neq 0$? Do not forget the exponents. The transformation to polar coordinates is $x = r \cos \theta$ and $y = r \sin \theta$, consequently

$$\frac{6x^3 y}{2x^4 + y^4} = \frac{6(r \cos \theta)^3 (r \sin \theta}{2(r \cos \theta)^4 + (r \sin \theta)^4} = \frac{r^4 6 \cos^3 \theta \sin \theta}{r^4 (2 \cos^4 \theta + \sin^4 \theta)} = \frac{6 \cos^3 \theta \sin \theta}{2 \cos^4 \theta+\sin^4 \theta}.$$

In my humble opinion your first approach is best and less confusing: two lines approaching the origin give two different limits and therefore it does not exist. Polar coordinates can be deceiving. What do you think will be the limit of

$$\lim_{(x,y) \to (0,0)} \frac{2x^2y}{x^4 +y^2}?$$
 

Petrus

Well-known member
Feb 21, 2013
739
Not really. As Petrus found, what we have is that

$$\lim_{r \to 0^+} \frac{6 \cos^3 \theta \sin \theta}{2 \cos^4 + \sin^4 \theta} = \frac{6 \cos^3 \theta \sin \theta}{2 \cos^4 + \sin^4 \theta},$$

since it is no longer dependent on $r$. The point is: this is now completely dependent on $\theta$. Therefore, taking $\theta = 0$ we get the zero answer while taking $\theta = \pi/4$ we get another nonzero answer. Hence the limit does not exist.


I do not follow your reasoning. Even though it will not exist, what does it have to do with the fact that $\sin \theta + \cos \theta \neq 0$? Do not forget the exponents. The transformation to polar coordinates is $x = r \cos \theta$ and $y = r \sin \theta$, consequently

$$\frac{6x^3 y}{2x^4 + y^4} = \frac{6(r \cos \theta)^3 (r \sin \theta}{2(r \cos \theta)^4 + (r \sin \theta)^4} = \frac{r^4 6 \cos^3 \theta \sin \theta}{r^4 (2 \cos^4 \theta + \sin^4 \theta)} = \frac{6 \cos^3 \theta \sin \theta}{2 \cos^4 \theta+\sin^4 \theta}.$$

In my humble opinion your first approach is best and less confusing: two lines approaching the origin give two different limits and therefore it does not exist. Polar coordinates can be deceiving. What do you think will be the limit of

$$\lim_{(x,y) \to (0,0)} \frac{2x^2y}{x^4 +y^2}?$$
Hello,
after some simplify I get
\(\displaystyle \lim_{r->0}\frac{2r\cos(\theta)\sin( \theta)}{r^2\cos^2(\theta)+\sin^2(\theta)}\)
the top Will always go to zero but the bottom Will go to zero if \(\displaystyle \theta=0\) so the limit does not exist?

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

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
You can't think as the quantities varying independently. Regardless of what the bottom goes, the whole fraction tends to zero. Considering the case $\theta = 0$ is simply saying $y=0$, which is in agreeing with the limit we have calculated.

However, consider the curve $y=x^2$. Approaching the origin with this gives us

$$\lim_{x \to 0} \frac{2x^2 (x^2)}{x^4 + (x^2)^2} = \lim_{x \to 0} \frac{2x^4}{2x^4} = 1.$$

Even though we have found the limit as zero when considering all straight lines through the origin, as we considered a parabola we've found a different value. Therefore, the limit does not exist.

This is a good example to illustrate two things: first, it is not enough to consider all straight lines; second, polar coordinates can deceive us. :)

Cheers!
 

Petrus

Well-known member
Feb 21, 2013
739
I just find something intressting I think..


When I calculate it and facit in My book says does not exist while WA says it should be zero..?

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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,776
I just find something intressting I think..

When I calculate it and facit in My book says does not exist while WA says it should be zero..?

Regards,
\(\displaystyle |\pi\rangle\)
Shows that W|A is a very helpful calculator.
As far as straight forward calculations go, you can trust it blindly.
But where special cases are concerned, you'd better be careful...