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Limits in polar coordinates

  • Thread starter quietrain
  • Start date
  • #1
654
2

Homework Statement


find limit if exist, or not then why

f(x,y) = xy / sqrt(x2+y2)
as (x,y) --> (0,0)

The Attempt at a Solution



can i change to polar coordinates
x=rcost
y=rsint

so f(r,t) = rcostsint

so as x,y tends to 0, r t will tend to 0 too?but cos(t) at 0 gives me 1?
so i can only rely on r?

so i say since r tends to 0 , rcostsint gives 0 , so the limit exist and is 0? but what about the variable t?

thanks!
 

Answers and Replies

  • #2
VietDao29
Homework Helper
1,423
2

Homework Statement


find limit if exist, or not then why

f(x,y) = xy / sqrt(x2+y2)
as (x,y) --> (0,0)

The Attempt at a Solution



can i change to polar coordinates
x=rcost
y=rsint

so f(r,t) = rcostsint

so as x,y tends to 0, r t will tend to 0 too?but cos(t) at 0 gives me 1?
When (x, y) tends to (0, 0), r will tend to 0, however t does not. This is because you can go to (0, 0) along many ways, so the angle t can be anything.

Say, if you go to (0, 0) along the x-axis from the left, then [tex]t = \pi[/tex].
If you go to (0, 0) along the y-axis from upwards, then [tex]t = \frac{\pi}{2}[/tex].

so i can only rely on r?

so i say since r tends to 0 , rcostsint gives 0 , so the limit exist and is 0? but what about the variable t?
If you can prove that the limit of your final expression (in r, and t) will tend to some value L as r tends to 0, and that limit does not depend on t. Then your original limit is also L.

In your example, we have: [tex]-r \le r\cos t\sin t \le r[/tex]. And let r tend to 0. By the Squeeze Theorem rcostsint will also tend to 0 (no matter what value of t is). So we have:

[tex]\lim_{(x, y) \rightarrow (0, 0)} \frac{xy}{\sqrt{x ^ 2 + y ^ 2}} = 0[/tex]

Is everything clear now? :)
 
  • #3
654
2
thank you so much!
 

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