Multivariable Calculus: Limits

[kingwinner]

1) lim [x(y^2)] / (x^2 + y^2)
(x,y)->(0,0)
Find the value of the given limit, if it exists.

Using polar coordinates, set x = r cos(theta), y = r sin(theta)
Then, the given limit = lim [r cos(theta) r^2 sin^2(theta)] / r^2
r->0
= lim r [cos(theta) sin^(theta)]
r->0
= 0 since cos(theta) sin^(theta)<=1, i.e. bounded

If I found that the limit is equal to 0 this way, can I conclude immediately that the original limit is 0 too?

What I believe is that for the part using polar coordinates r->0, it seems that it's appraoching the origin through straight lines paths ONLY, however my textbook says that "for the limit to exist, we must get the same result no matter which of the infinite number of paths is chosen"


Thanks for answering!

 

[b0it0i]

If I found that the limit is equal to 0 this way, can I conclude immediately that the original limit is 0 too?

yes you cani believe when you switch to polarand after simplification

plug in your r

if your equation is not a function of theta, then that is the limitif your equation is still a function of theta, then the limit DNE
and you can also use lhopitals rule once you switch to polar
if the conditions are met

 

[HallsofIvy]

What I believe is that for the part using polar coordinates r->0, it seems that it's appraoching the origin through straight lines paths ONLY, however my textbook says that "for the limit to exist, we must get the same result no matter which of the infinite number of paths is chosen"


Thanks for answering!

No, you are not approaching "through straight lines only" because you are not assuming that \theta is a constant. You need to show that the limit goes to 0 as r goes to 0, no matter what \theta is. The real point is that, in polar coordinates, r measures exactly the distance to the origin, irrespective of \theta- and that is what must be made "less than \delta".

 

[kingwinner]

"You need to show that the limit goes to 0 as r goes to 0, no matter what theta is" <---but this covers only the paths of all the straight lines through the origin, how about in the paths of parabolas and cube root function through the origin, etc?

 

[genneth]

Because theta can be a function of r...

 

[kingwinner]

Because theta can be a function of r...

But how does this help? Sorry, I am not getting it...

 

[genneth]

It's a straight line iff theta is a constant. But the proof allows theta to vary arbitrarily, therefore, you're allowed curves.

 

[kingwinner]

How can it be a curve if theta can vary?

 

[genneth]

Consider the equation \theta = r. Plot it. Is it not a curve?

 

[arildno]

How can it be a curve if theta can vary?

It seems you are unsure of what theta really is.

So I have to ask:

What, precisely, does the variable theta measure?

 

[kingwinner]

It seems you are unsure of what theta really is.

So I have to ask:

What, precisely, does the variable theta measure?

theta is a counterclockwise measure of "angle" from the positive x-axis, this is what makes me think of r->0 as (x,y) approaching (0,0) through straight lines only

 

[genneth]

Did you try and plot the function \theta = r as I suggested? It is a straight line?

 

[kingwinner]

Did you try and plot the function \theta = r as I suggested? It is a straight line?

Do you mean [r cos(r) r^2 sin^2(r)] / r^2 ?

By the way, we are taking the limit r->0, shouldn't theta be fixed?

 

[arildno]

theta is a counterclockwise measure of "angle" from the positive x-axis, this is what makes me think of r->0 as (x,y) approaching (0,0) through straight lines only

Why?
Any particular POINT can be uniquely specified by its distance from the origin, and the angle the line segment between the origin and the point makes with the positive x-axis.

By no means does this entail that all the points a particular curve consists of make the same angle to the positive x-axis with their respective line segments.

 

[genneth]

By the way, we are taking the limit r->0, shouldn't theta be fixed?

No! That's the point! We DON'T have to fix theta!

 

[arildno]

You can approach the origin by way of a straight line, but also by spiralling yourself inwards. And in many other ways as well.
Only if you approach the originalong a straight line will theta be constant.

 

[kingwinner]

If theta is allowed to vary, will r->0 using polar coordinates cover ALL paths for which (x,y)->(0,0) ?

 

[genneth]

Yes.

 

[HallsofIvy]

"Paths", in polar coordinates, depend only on r and \theta. If r ranges from, say, 1 to 0, while \theta is allowed to have any value, then, yes, it covers all paths. More to the point is that in polar coordinates, r alone measures the distance from the origin which is what you want: ||(x,y)- (0,0)||= r< \delta.
As long as the limit, as r goes to 0, is a number, that is, independent of \theta that number will be the limit.

 

[genneth]

I think part of the confusion might be this: a path near the origin, if it goes through it, can be well approximated by a straight line. Unfortunately, whilst that's intuitively true, analysis can always come back to bite us:

\theta = \pi sin(1/r)

 

[kingwinner]

1) Using polar coordinates, how can it cover the paths of say, parabolas, in approaching (0,0) ? (if theta can vary)

Say, if I am trying to eavluate a limit by changing it to polar coordinates, and I get a finite limit L using the polar coordinates, can I ALWAYS immediately conclude that the original limit is L in any case like this?

 

[genneth]

Let's just get things clear. The definition of limits require a metric space. In R^n, there's the obvious euclidean metric. That means when you evaluate it, you convert to a "polar" representation (or more accurately, hyperspherical). However, that's not the only one possible -- you've got the infinite family of L^n norms, and that's just in R^n.

So, in R^2, in the case that you're using the euclidean norm (as you are), the definition of limit requires that you convert to polar coordinates.

The question of finding the polar representation of y=x^2 is left as an exercise.

 

[HallsofIvy]

1) Using polar coordinates, how can it cover the paths of say, parabolas, in approaching (0,0) ? (if theta can vary)

If, say, the parabola is y= ax², replacing y by r sin(\theta) and x by r sin(\theta), then the parabola becomes r sin(\theta)= r^2 cos^2(\theta) or r= tan(\theta)sec(\theta). Now, with that relationship between r and \theta, as r goes to infinity, the point (x,y) will go to (0,0) along that curve. But, of course, the whole point is to allow r itself to go to 0, without regard for what \theta is.

Say, if I am trying to eavluate a limit by changing it to polar coordinates, and I get a finite limit L using the polar coordinates, can I ALWAYS immediately conclude that the original limit is L in any case like this?

Yes, IF that finite limit is as r goes to 0 and does NOT depend on \theta.

For example, if f is any function, r f(\theta)+ 5, as r goes to 0, goes to 5 whatever \theta is- if you had a function that reduced to that in polar coordinates, its limit at (0,0) would be "5". But r+ 5f(\theta), as r goes to 0, goes to 5f(\theta) and so the value very close to (0,0) depends upon \theta and so a function that reduced to that in polar coordinates would NOT have a limit at (0,0).