Question

how can i solve " lim x^m y^n /x^2+2kxy+y^2" x,y approaches to (0,0)

Answer 1

what is k=..?

Answer 2

tryin to think of a case where limit will be different if u approach along two functions..

Answer 3

|k|<1

Answer 4

and m+n>2

Answer 5

you know what.. I really think this depends on whether either m or n are odd or even..

Answer 6

Here's what I mean.. suppose m is odd, and suppose we approach along x = -1, then you will have x^m, which will be -1^m, and if m is odd, then you will have -y^n/1-2ky+y^2.. wait a second.. that's 0... hmm weird, I started to think that this limit exists...

Answer 7

just a thought.....:
\[=\lim[ x^{m}*y ^{n}/((x+y)^{2}+2xy(k-1))]=\]
if |k|<1, then if x, y >0
(x+y)^2 + 2xy(k-1) <=((x+y)^2
then:
\[\lim <=limx ^{m}y ^{n}/(x+y)^{2}\]

Am I right so far...?

Answer 8

inik is the math showing up properly on your pc?

Answer 9

who knows...
what I see:
\[\lim x ^{m}y ^{n}/(x ^{2}+2kxy + y ^{2})\]

is it wrong?

Answer 10

no,it is true

Answer 11

oh wait it was a problem with my browser, but anyway when u said:
then if x, y >0
(x+y)^2 + 2xy(k-1) <=((x+y)^2, why did u plug in 0s into 2xy but not into x+y?

Answer 12

actually I'm looking Stewart Calculus now on this matter.
it seems much easier solution:

Answer 13

wish i had that book.. =(

Answer 14

when function approach (0,0) along the x-axis, then y=0 gives f(x,0)=0/x^2 = 0 for all x not equal 0,
so f(x,y) ->0 as (x,y) -> (0,0) along the x-axis

the same - for y-axis

Answer 15

when function approach (0,0) along the y-axis, then x=0 gives f(x,0)=0/y^2 = 0 for all y not equal 0,
so f(x,y) ->0 as (x,y) -> (0,0) along the y-axis

therefore f(x,y) ->0 as (x,y)->(0,0)

do you agree?