Question

Another limit question

Answer 1

\[
\lim_{(h,k)\to (0,0)}\frac{\bigg(\bigg(\sqrt{(x+h)^2+y^2}\bigg)^{n-1} x+h\bigg )^n- \bigg (\bigg(\sqrt{x^2+y^2}\bigg)^{n-1} x\bigg)^n}{\sqrt{h^2+k^2}}
\]

Answer 2

Okay, I know this one is really nasty, I apologize.
I think we can consider different cases for \(n\) though.

Answer 3

Also \(x,y\) are constants, so don't worry about them changing with respect to \(h,k\).

Answer 4

I can introduce the origin of this limit if you're interested.

Answer 5

na , this is just a f'(x)

Answer 6

for cute f(x) function

Answer 7

oh . sorry i mean f(x,y)

Answer 8

Yes, it's origin comes from differentiation, but it's not that simple, in my opinion.

Answer 9

Marki is correct,
also you can simply put in k = 0
and make it limit with h->0 only

Answer 10

\(\lim_{(h,k)\to (0,0)}\frac{\bigg(\bigg(\sqrt{(x+h)^2+y^2}\bigg)^{n-1} (x+h)\bigg )^n- \bigg (\bigg(\sqrt{x^2+y^2}\bigg)^{n-1} x\bigg)^n}{\sqrt{h^2+k^2}}\)

missed the brackets around x+h ?

Answer 11

Okay, sure, but here is the problem, you let \(k\to0\) and then you have: \[
\frac{\ldots }{|h|}
\]

Answer 12

And last limit sort of showed how \(\frac xx\) and \(\frac x{|x|}\) are different

Answer 13

There are no brackets around \(x+h\)

Answer 14

please post the source of this question too :)

Answer 15

if there are no brackets around x+h
there the numerator is NOT of the form
f(x+h) -f(x)

Answer 16

let \( f(x,y)=\bigg (\bigg(\sqrt{x^2+y^2}\bigg)^{n-1} x\bigg)^n \)
then ur limit is the same as
\( f(x,y)' \)

Answer 17

hmm abt |h| and |k|
lets see if it make sense

Answer 18

Okay, it's something I came up on my own. Let \[
\mathbf u\mathbf v = \|\mathbf v\|\mathbf u
\]And let: \[
\frac{\mathbf u}{\mathbf v} = \frac{\mathbf u}{\|\mathbf v\|}
\]I'm guessing this would mean: \[
\mathbf u^n = \|\mathbf u \|^{n-1}\mathbf u
\]

Answer 19

I let \(\mathbf u=\langle x,y\rangle\), then I let \(\mathbf v=\langle h,k\rangle\).
Then I tried to do: \[
\lim_{\mathbf v\to(0,0)}\frac{(\mathbf u+\mathbf v)^n-\mathbf u^n}{\mathbf v}
\]I only focused on the \(x\) part though. Maybe there should be parenthesis around \(x+h\) though...

Answer 20

Yes, put parenthesis around the \(x+h\) in there.

Answer 21

hmm as u define the denominator it in ur limits , u define a disk or a ball or name it wat u like , lets say " singular point " , thats why its needed and its not the same as |h|
in x limit

Answer 22

What?

Answer 23

>.<

Answer 24

consider two paths
h = 0 and k = 0

and show that the limit evaluates to different values

Answer 25

Is there a way to make the limit exist?

Answer 26

Like an alteration to it that would make it work?

Answer 27

you're trying to make the limit exist by modifying the function slightly ?

Answer 28

This limit has the same problem that: \[
\frac{x}{|x|} = \begin{cases}
-1&x<0\\
1&x>0
\end{cases}
\]has, but you can modify \(x/|x|\) somehow to make it \(x/x\) or \(|x|/|x|\) and it should exist.

Answer 29

@ganeshie8 Yeah, I want to modify the function slightly... actually maybe my \(\circ\) operation is what needs to be modified.

Answer 30

Kinda getting... need to study this more... but as it stands the limit DNE as you can find two different paths evaluating to different values

Answer 31

If I let \(n=1\), I get a simple limit: \[
\lim_{(h,k)\to (0,0)}\frac{h}{\sqrt{h^2+k^2}} = \lim_{(h,k)\to (0,0)}\frac{h}{|h|\sqrt{1+(k/h)^2}}
\]

Answer 32

I know that this doesn't exist for all paths, I'm wondering if there is slight modification that would make it work somehow.

Answer 33

Now that I think about it, I don't think there is anything I can do here. If \(k=mh\), then it willl always depend on what \(m\) is... damn!

Answer 34

By the way... if you let \[
\mathbf u\circ \mathbf v = \langle u_1v_1-u_2v_1,u_2v_1+u_1v_2\rangle
\]Then it becomes the complex derivative definition.

Answer 35

But that assumes \(\mathbf u \in\mathbb R^2\).

Answer 36

When you let \[
\mathbf u\circ \mathbf v = \langle u_1v_1,u_2v_2,\ldots ,u_nv_n\rangle
\]It seems to become a sort of gradient operation.

Answer 37

I have another limit question