Question

Curious about this limit

Answer 1

\[
\lim_{(h,k)\to(0,0)}\frac{\langle (x+h)^n-x^n,(y+k)^n-y^n\rangle}{\sqrt{h^2+k^2}}
\]

Answer 2

We can split it up into two completely separate and symmetric limits though.

Answer 3

\[
\lim_{(h,k)\to (0,0)}\frac{(x+h)^n-x^n}{\sqrt{h^2+k^2}}
\]

Answer 4

I haven't learned this yet. What's a limit.

Answer 5

In short, a limit tells you what a function's output has to be for it to be continuous.

Answer 6

Oh. Just found out a bit. It's calculus and I haven't reached that yet.

Answer 7

@zepdrix @satellite73 Do you guys have any ideas on this?

Answer 8

Perhaps the limit doesn't exist if I let \(k=mh\)

Answer 9

@zzr0ck3r What do you think?

Answer 10

Am I wrong in thinking you just plug in zero. So in this case you'd get 0/0 so does not exist?

Answer 11

Okay, so for \(n=0\), the limit clearly goes to \(0\).

Answer 12

Isn't h and k going to zero?

Answer 13

Yeah, but what I mean is that when \(n=0\), the whole thing becomes the constant \(0\), which is why the limit goes to \(0\).

Answer 14

When \(n=1\), we get: \[
\lim_{(h,k)\to (0,0)}\frac{h}{\sqrt{h^2+k^2}}
\]

Answer 15

hehe I would put it in a calculator:) I am not so hot at finding limits. But if you find it, we can try and prove it:)

Answer 16

You still have zero in the denominator in both cases

Answer 17

Letting \(k=mh\), then: \[
\frac{h}{\sqrt{h^2+m^2h^2}} = \frac{h}{|h|\sqrt{1+m^2}}
\]I think this limit doesn't exist

Answer 18

@ChmE You wouldn't have a \(0\) in the denominator at any point that isn't \((0,0)\), yet you would get a value of \(0\) in the numerator, so that is why it would have to be \(0\) when \(n=0\).

Answer 19

Correct. I was just looking at it in terms of the limit. So only at (0,0)

Answer 20

I think you've thought it through nicely with your last derivation