anonymous
  • anonymous
Find the value of k such that f(x,y) is continuous at (0,0). Is f(x,y) then continuous everywhere? why?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1378590899365:dw|
anonymous
  • anonymous
So I basically get: \[\lim_{(x,y) \rightarrow (0,0)}\frac{ x^4 }{ x^2+y^2 }=\lim_{(x,y) \rightarrow (0,0)}k\]
anonymous
  • anonymous
Could I convert the lhs to polar coordinates afterwards?

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anonymous
  • anonymous
So it would just imply k=o wouldn't it I guess?
blockcolder
  • blockcolder
Have you examined the limit from a variety of directions?
anonymous
  • anonymous
That does not prove the limit exists. @blockcolder
anonymous
  • anonymous
@zepdrix @ganeshie8
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
@wio
anonymous
  • anonymous
The limit would have to exist for \(k\) to exist.
anonymous
  • anonymous
What if you approach from \(y=x\)?
anonymous
  • anonymous
Well the thing is I can approach a limit from many point but unless I test an infinite number of points the limit cannot be shown to exist.
anonymous
  • anonymous
paths not points sorry.
anonymous
  • anonymous
I should mention I solved for the limit already. It's 0 :P .
anonymous
  • anonymous
The only way to be certain is to use the definition of the limit.
anonymous
  • anonymous
But that is not what they expect you to do here.
anonymous
  • anonymous
Well if I get that the LHS is 0 does that mean k =0 as well?
anonymous
  • anonymous
|dw:1378606511227:dw|
anonymous
  • anonymous
I'm not sure what you mean.
anonymous
  • anonymous
Well continuity implies that the LHS limit and the RHS limit must be equal right?
anonymous
  • anonymous
Continuity implies the limit exists period. LHS and RHS don't apply to multivariable limits.
anonymous
  • anonymous
Yeah. I converted to polar coordinates for that.
anonymous
  • anonymous
I have a question. When doing multivariable limits, can't you change them into iterated limits?
anonymous
  • anonymous
Like x=0 and evaluate the limit? Like that?
anonymous
  • anonymous
Like the path y=x or y=0, x=0? Like that?
anonymous
  • anonymous
Apparently you can't do that: http://en.wikipedia.org/wiki/Iterated_limits_of_functions_of_two_variables
anonymous
  • anonymous
Yeah. You have to approach it from a specific path. The iterated idea only works for integrals.
anonymous
  • anonymous
I don't know any rules for multivariable limits that would make this doable without using the definition of a limit.
anonymous
  • anonymous
I didn't use the epsilon delta definition for limits though.
anonymous
  • anonymous
Couldn't I convert it to polar coordinates in order to make this single variable?
anonymous
  • anonymous
Yeah
anonymous
  • anonymous
Well then I just get: \[\lim_{r \rightarrow 0}r^2\cos^4(\theta)\]
anonymous
  • anonymous
And that's 0 by the squeeze theorem.
anonymous
  • anonymous
Squeeze theorem? You're not squeezing it though, lol.
anonymous
  • anonymous
Well my prof always did it for some reason :P . It's still 0 haha.
anonymous
  • anonymous
Okay okay :P . Still 0 though.
anonymous
  • anonymous
Squeeze theorem would be like: \[ \frac{0}{x^2+y^2}\leq \frac{x^4}{x^2+y^2} \leq \frac{x^2}{x^2+y^2} \]
anonymous
  • anonymous
So it's basically saying: |dw:1378607532358:dw|
anonymous
  • anonymous
But yeah, polar coords should work, I guess.
anonymous
  • anonymous
Why didn't they work before though?
anonymous
  • anonymous
No I am just asking :P . Is converting correct?
anonymous
  • anonymous
You don't need to do \(\lim k\) here.
anonymous
  • anonymous
Because I have no idea how to do it otherwise without using Cauchy-Shawrz.
anonymous
  • anonymous
ohh?
anonymous
  • anonymous
You just need to say \[ \lim_{(x,y)\to (0.0)}f(x,y) = f(0,0) \]
anonymous
  • anonymous
This is the definition of continuity at the point \((0,0)\).
anonymous
  • anonymous
You're rigging it so that this is the case.
anonymous
  • anonymous
It's been so long since I took Calc I >.<. yeah that works :) .
anonymous
  • anonymous
But wait. That just means that the limit would equal k.
anonymous
  • anonymous
|dw:1378607865172:dw|
anonymous
  • anonymous
|dw:1378607897982:dw|
anonymous
  • anonymous
|dw:1378607932307:dw|
anonymous
  • anonymous
|dw:1378607952923:dw|
anonymous
  • anonymous
Right?
anonymous
  • anonymous
Yeah
anonymous
  • anonymous
Thank you!! :) .
anonymous
  • anonymous
I just realized something...
anonymous
  • anonymous
?
anonymous
  • anonymous
The limit before that didn't exist... When we showed that it evaluated to \(\cos^2(\theta)-\sin^2(\theta)\) we kinda had proven the limit didn't exist because it was \(\theta\) (direction) dependent. Interesting.
anonymous
  • anonymous
yeah I solved that using direct substitution before :) .
anonymous
  • anonymous
I tried y=x and y=0. The two limits differ so the limit DNE.
anonymous
  • anonymous
Yeah, and \(y=0\) is equivalent to saying \(\theta = 0\). \(y=x\) is equivalent to \(\theta = \pi / 4\)
anonymous
  • anonymous
Okay :) . That works too.

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