Find the value of k such that f(x,y) is continuous at (0,0). Is f(x,y) then continuous everywhere? why?

- anonymous

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- anonymous

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- 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

Could I convert the lhs to polar coordinates afterwards?

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## More answers

- anonymous

So it would just imply k=o wouldn't it I guess?

- blockcolder

Have you examined the limit from a variety of directions?

- anonymous

That does not prove the limit exists. @blockcolder

- anonymous

@zepdrix @ganeshie8

- anonymous

@dan815

- anonymous

@wio

- anonymous

The limit would have to exist for \(k\) to exist.

- anonymous

What if you approach from \(y=x\)?

- 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

paths not points sorry.

- anonymous

I should mention I solved for the limit already. It's 0 :P .

- anonymous

The only way to be certain is to use the definition of the limit.

- anonymous

But that is not what they expect you to do here.

- anonymous

Well if I get that the LHS is 0 does that mean k =0 as well?

- anonymous

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- anonymous

I'm not sure what you mean.

- anonymous

Well continuity implies that the LHS limit and the RHS limit must be equal right?

- anonymous

Continuity implies the limit exists period. LHS and RHS don't apply to multivariable limits.

- anonymous

Yeah. I converted to polar coordinates for that.

- anonymous

I have a question. When doing multivariable limits, can't you change them into iterated limits?

- anonymous

Like x=0 and evaluate the limit? Like that?

- anonymous

Like the path y=x or y=0, x=0? Like that?

- anonymous

Apparently you can't do that:
http://en.wikipedia.org/wiki/Iterated_limits_of_functions_of_two_variables

- anonymous

Yeah. You have to approach it from a specific path. The iterated idea only works for integrals.

- anonymous

I don't know any rules for multivariable limits that would make this doable without using the definition of a limit.

- anonymous

I didn't use the epsilon delta definition for limits though.

- anonymous

Couldn't I convert it to polar coordinates in order to make this single variable?

- anonymous

Yeah

- anonymous

Well then I just get:
\[\lim_{r \rightarrow 0}r^2\cos^4(\theta)\]

- anonymous

And that's 0 by the squeeze theorem.

- anonymous

Squeeze theorem? You're not squeezing it though, lol.

- anonymous

Well my prof always did it for some reason :P . It's still 0 haha.

- anonymous

Okay okay :P . Still 0 though.

- 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

So it's basically saying:
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- anonymous

But yeah, polar coords should work, I guess.

- anonymous

Why didn't they work before though?

- anonymous

No I am just asking :P . Is converting correct?

- anonymous

You don't need to do \(\lim k\) here.

- anonymous

Because I have no idea how to do it otherwise without using Cauchy-Shawrz.

- anonymous

ohh?

- anonymous

You just need to say \[
\lim_{(x,y)\to (0.0)}f(x,y) = f(0,0)
\]

- anonymous

This is the definition of continuity at the point \((0,0)\).

- anonymous

You're rigging it so that this is the case.

- anonymous

It's been so long since I took Calc I >.<. yeah that works :) .

- anonymous

But wait. That just means that the limit would equal k.

- anonymous

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- anonymous

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- anonymous

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- anonymous

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- anonymous

Right?

- anonymous

Yeah

- anonymous

Thank you!! :) .

- anonymous

I just realized something...

- 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

yeah I solved that using direct substitution before :) .

- anonymous

I tried y=x and y=0. The two limits differ so the limit DNE.

- anonymous

Yeah, and \(y=0\) is equivalent to saying \(\theta = 0\).
\(y=x\) is equivalent to \(\theta = \pi / 4\)

- anonymous

Okay :) . That works too.

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