# Limit in two variables polar vs cartisiean differ in result

1. Nov 5, 2012

### ahmeeeeeeeeee

Hello

I have the limit

lim (x^9 * y) / (x^6 + y^2)^2

(x,y)---> (0,0)

when I use polar the final result is

limit =

lim (r^6 cos^9 (theta) sin (theta) ) / (r^4 cos^6 (theta) + sin^2 (theta))

r--->0

and substituting r = 0 , it will give zero

* I tried it on wolfram alpha and it gave zero

http://www.wolframal...%3E+%280%2C0%29 [Broken]

but when I use cartezean and try the path y = mx^3

the result turns to be
m/(1+m^2)^2

which depends on m

so what is it ?!

I may think that the limit

(r^6 cos^9 (theta) sin (theta) ) / (r^4 cos^6 (theta) + sin^2 (theta))

has a problem of the sin^2 theta downwards , as it may become zero and the limit won't be zero / positive value any more

but then it is 0/0 , so what to do next to know whether this (x/x) will turn eventually to be zero in the limit or not ?!

Last edited by a moderator: May 6, 2017
2. Nov 5, 2012

### arildno

Very interesting!
My initial reaction is that you have found a bug in Wolfram (good feat! )
As for the coordinate transformations, they don't matter, since the actual sine values (in "y") will decrease along the powers of "r" in perfectly acceptable fashion (namely, proportional to r^2) when following your chosen curve.
I've notified the others Advisors on your conundrum as being of interest.

3. Nov 5, 2012

### DonAntonio

You seem to have commited several mistakes (which, btw, is another good reason why it is higly advisable to use LaTeX to write in this site).

Using $\,x=r\cos t\,\,,\,\,y=r\sin t\,$ , we get:

$$\frac{x^9y}{(x^6+y^2)^2}=\frac{r^{10}\cos t\sin t}{r^4(r^4\cos^6 t+\sin^2t)^2}=r^6\frac{\cos t\sin t}{(r^4\cos^6 t+\sin^2t)^2}\xrightarrow [r\to 0]{} \text{doesn't exist}$$

If you use $\,y=mx^3\,$ nothing changes:

$$\frac{x^9y}{(x^6+y^2)^2}=\frac{mx^{12}}{(x^6+m^2x^6)^2}\xrightarrow [x\to 0]{}\frac{m}{(1+m^2)^2}$$

so the limit depends on $\,m\,$ and thus doesn't exist.

DonAntonio

Last edited by a moderator: May 6, 2017
4. Nov 5, 2012

### arildno

The question, DonAntonio, is whether WOLFRAM is making those mistakes, not the OP.
The limit doesn't exist (OPs example versus, say, x=0), but according to OP, Wolfram says it does.

5. Nov 5, 2012

### AlephZero

Is Wolfram taking the limits one at a time?
$$\lim_{x\rightarrow 0}\left(\lim_{x \ne 0, y\rightarrow 0} \frac{x^9 y} {(x^6 + y^2)^2}\right) = \lim_{x\rightarrow 0}(0) = 0$$
and also
$$\lim_{y\rightarrow 0}\left(\lim_{y \ne 0, x\rightarrow 0} \frac{x^9 y} {(x^6 + y^2)^2}\right) = 0$$

6. Nov 7, 2012

### ahmeeeeeeeeee

Very interesting!
My initial reaction is that you have found a bug in Wolfram (good feat! )

:)

Thanks a lot all , It's not good to lose confidence in wolfram :)

would they give me a free membership If I send them this mistake ? :D

well actually I do want to get premium , but I don't know how to pay for it , here in Egypt . Is there any bank account to send money to instead of credit cards ?!

as to

$\frac{r^6 cosθ sinθ}{(r^4 cos^6 θ +sin^2 θ)^2}$

why doesn't it exist ?!

I see it always equals zero , unless the " sin^2 θ " equals zero then the result will be $\frac{0}{0}$ which is unknown value , not doesn't exist , so if we find that the limit ( by some way I am asking for if there is one) also equals zero , then the limit is zero

( for this example I think we now that the limit when sinθ =zero will not become zero and so the limit doesn't exist , but I think we know that because we tried the other method, so I am asking is there something else we could do ?)

thanks

7. Nov 7, 2012

### arildno

If you let "theta" have a FIXED, non-zero value, that is traversing along straight rays towards the origin, then the limit IS 0.

But y=mx^3 says that sin(theta)=mr^2cos^3(theta) along THAT curve.

Last edited: Nov 7, 2012
8. Nov 7, 2012

### ahmeeeeeeeeee

IF another limit turns to be

limit (z) = r cos(2t)sin(t)
r->0

for example , wouldn't this make the limit value =zero , and wouldn't this make it impossible for any other curve to make the limit dependent on another parameter because I didn't choose a certain curve in polar ?! or am I understanding wrong ?!

secondly

I took the result from above and put it in wolfram alpha again

(( because if the problem in the first result was that when theta equals zero the limit will be 0/0 , I took the limit when r and theta both equal zero )) and it gave zero again

(would you see this)
http://www.wolframalpha.com/input/?i=limit+%28r^6+cos^9+%28t%29+sin+%28t%29+%29+%2F+%28r^4+cos^6+%28t%29+%2B+sin^2+%28t%29%29+as+%28r%2Ct%29--%3E%280%2C0%29

I now want to know something

is the limit :-

$\frac{r^6 cosθ sinθ}{(r^4 cos^6 θ +sin^2 θ)^2}$

when both r and θ tend to zero

1- equal to zero really and despite that, the original limit doesn't exist ( and so polar isn't general )

OR

2- it isn't equal to zero and there is a problem with the programming of wolfram alpha ( and so polar is general)

9. Nov 7, 2012

### arildno

"
for example , wouldn't this make the limit value =zero , and wouldn't this make it impossible for any other curve to make the limit dependent on another parameter because I didn't choose a certain curve in polar ?! or am I understanding wrong ?!"

No, you are quite right.
the ONLY way such expressions could give another result is if one of the trigonemtric factors behaved, for example, as 1/r as r tends to zero, but that is impossible, since the absolute value of sine and cosine can never exceed 1.

10. Nov 7, 2012

### arildno

"when both r and θ tend to zero "
This is NOT right!
You can close in on the origin from EVERY angle, and thus "theta going to zero" would be to merely evaluate along those curves closing in on the x-axis when approaching 0.

The limit in polar coordinates must hold for ALL legitimate theta-choices in order to be said to exist.

11. Nov 7, 2012

### ahmeeeeeeeeee

""theta going to zero" would be to merely evaluate along those curves closing in on the x-axis when approaching 0.
""

ok, thank you , this note wasn't in my mind :)

12. Nov 7, 2012

### arildno

It might be you gave WA a wrong command, by giving it a requirement of theta=constant.

But, from what I have heard, WA is not wholly fool-proof in evaluating multivariable limits properly.